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## Section6.1Using Definite Integrals to Find Area and Length

###### Motivating Questions
• How can we use definite integrals to measure the area between two curves?

• How do we decide whether to integrate with respect to $x$ or with respect to $y$ when we try to find the area of a region?

• How can a definite integral be used to measure the length of a curve?

Early on in our work with the definite integral, we learned that for an object moving along an axis, the area under a non-negative velocity function $v$ between $a$ and $b$ tells us the distance the object traveled on that time interval, and that area is given precisely by the definite integral $\int_a^b v(t) \, dt\text{.}$ In general, for any nonnegative function $f$ on an interval $[a,b]\text{,}$ $\int_a^b f(x) \, dx$ measures the area bounded by the curve and the $x$-axis between $x = a$ and $x = b\text{.}$

Next, we will explore how definite integrals can be used to represent other physically important properties. In Example6.1, we investigate how a single definite integral may be used to represent the area between two curves.

###### Example6.1

Consider the functions given by $f(x) = 5-(x-1)^2$ and $g(x) = 4-x\text{.}$

1. Use algebra to find the points where the graphs of $f$ and $g$ intersect.

2. Sketch an accurate graph of $f$ and $g$ on the axes provided, labeling the curves by name and the intersection points with ordered pairs.

3. Find and evaluate exactly an integral expression that represents the area between $y = f(x)$ and the $x$-axis on the interval between the intersection points of $f$ and $g\text{.}$

4. Find and evaluate exactly an integral expression that represents the area between $y = g(x)$ and the $x$-axis on the interval between the intersection points of $f$ and $g\text{.}$

5. What is the exact area between $f$ and $g$ between their intersection points? Why? Figure6.2Axes for plotting $f$ and $g$ in Example6.1
Hint
1. Solve the quadratic equation $f(x)=g(x)\text{.}$
2. The graph of $f(x)$ is a parabola, and the graph of $g(x)$ is a line.
3. The area will be $\int_a^b f(x)\, dx$ where $a$ and $b$ are the $x$-coordinates you found in part (a.).
4. The area will be $\int_a^b g(x)\, dx$ where $a$ and $b$ are the $x$-coordinates you found in part (a.).
5. You can find the area between them by subtracting your answer to the previous parts.
1. The intersection points are $(0,4)$ and $(3,1)\text{.}$
2. 3. 12
4. 7.5
5. 4.5
Solution
1. We set $5-(x-1)^2= 4-x$ and solve to get that $0=x^2-3x = x(x-3)\text{,}$ so the intersection points are at $x=0$ and $x=3 \text{.}$ Plugging these into $f(x)$ or $g(x) \text{,}$ we get that the intersection points are $(0,4)$ and $(3,1)\text{.}$
2. 3. The area between a curve and the $x$-axis is the integral of the function between the two $x$-coordinates. Thus our answer is
\begin{equation*} \int_0^3 f(x)\, dx = \int_0^3 5-(x-1)^2 \, dx = \left. 5x - \frac 13 (x-1)^3 \right |_0^3 \\ =\left(15-\frac 83 \right) - \left(0 +\frac 13 \right) = 12 \end{equation*}
4. The area between a curve and the $x$-axis is the integral of the function between the two $x$-coordinates. Thus our answer is
\begin{equation*} \int_0^3 g(x)\, dx = \int_0^3 4-x\, dx = \left. 4x - \frac 12 x^2 \right |_0^3 \\ =\left(12-\frac 92 \right) - \left(0 -0 \right) = 7.5 \end{equation*}
5. The area between $f(x)$ and $g(x)$ is all of the area below $f(x)\text{,}$ but not below $g(x)\text{.}$ To get this, we must take all of the area below $f(x)$ (our answer to part (c.)) and subtract all of the area below $g(x)$ (our answer to part (d.)). That is, our answer is
\begin{equation*} \int_0^3 f(x)\, dx - \int_0^3 g(x)\, dx = 12-7.5 = 4.5 \end{equation*}

### SubsectionThe Area Between Two Curves

In Example6.1, we saw a natural way to think about the area between two curves: it is the area beneath the upper curve minus the area below the lower curve.

###### Example6.3

Find the area bounded between the graphs of $f(x) = (x-1)^2 + 1$ and $g(x) = x+2\text{.}$ Figure6.4The first two graphs show the area under the curve $g(x)$ and $f(x) \text{,}$ respectively, on the interval $[0,3]\text{.}$ The third graph shows the area bounded by the functions $f(x) = (x-1)^2 + 1$ and $g(x) = x+2$ on the interval $[0,3]\text{.}$
Solution

In Figure6.4, we see that the graphs intersect at $(0,2)$ and $(3,5)\text{.}$ We can find these intersection points algebraically by solving the system of equations given by $y = x+2$ and $y = (x-1)^2 + 1\text{:}$ substituting $x+2$ for $y$ in the second equation yields $x+2 = (x-1)^2 + 1\text{,}$ so $x+2 = x^2 - 2x + 1 + 1\text{,}$ and thus

\begin{equation*} x^2 - 3x = x(x-3) = 0\text{,} \end{equation*}

from which it follows that $x = 0$ or $x = 3\text{.}$ Using $y = x+2\text{,}$ we find the corresponding $y$-values of the intersection points.

On the interval $[0,3]\text{,}$ the area beneath $g$ is

\begin{equation*} \int_0^3 (x+2) \, dx = \frac{21}{2}\text{,} \end{equation*}

while the area under $f$ on the same interval is

\begin{equation*} \int_0^3 [(x-1)^2 + 1] \, dx = 6\text{.} \end{equation*}

Thus, the area between the curves is

\begin{equation} A = \int_0^3 (x+2) \, dx - \int_0^3 [(x-1)^2 + 1] \, dx = \frac{21}{2} - 6 = \frac{9}{2}\text{.}\label{E-DiffOfInt}\tag{6.1} \end{equation}

We can also think of the area this way: if we slice up the region between two curves into thin vertical rectangles (in the same spirit as we originally sliced the region between a single curve and the $x$-axis in Section4.2), we see (as shown in Figure6.5) that the height of a typical rectangle is given by the difference between the two functions, $g(x) - f(x)\text{,}$ and its width is $\Delta x\text{.}$ Thus the area of the rectangle is

\begin{equation*} A_{\text{rect} } = (g(x) - f(x)) \Delta x\text{.} \end{equation*} Figure6.5On the left is an example of a rectangular vertical slice. The right figure shows the area bounded by the functions $f(x) = (x-1)^2 + 1$ and $g(x) = x+2$ on the interval $[0,3]\text{.}$

The area between the two curves on $[0,3]$ is thus approximated by the Riemann sum

\begin{equation*} A \approx \sum_{i=1}^{n} (g(x_i) - f(x_i)) \Delta x\text{,} \end{equation*}

and as we let $n \to \infty\text{,}$ it follows that the area is given by the single definite integral

\begin{equation} A = \int_0^3 (g(x) - f(x)) \, dx\text{.}\label{E-IntOfDiff}\tag{6.2} \end{equation}

We will see this strategy many times in the following sections: take a representative slice, approximate its area in terms of $\Delta x\text{,}$ sum them up, and then take a limit. When we do so, the sum will change to an integral and $\Delta x$ will turn into a $dx\text{.}$ Here, the integral sums the areas of thin rectangles.

Finally, it doesn't matter whether we think of the area between two curves as the difference between the area bounded by the individual curves (as in(6.1)) or as the limit of a Riemann sum of the areas of thin rectangles between the curves (as in(6.2)). These two results are the same, since the difference of two integrals is the integral of the difference:

\begin{equation*} \int_0^3 g(x) \, dx - \int_0^3 f(x) \, dx = \int_0^3 (g(x) - f(x)) \, dx\text{.} \end{equation*}

Our work so far in this section illustrates the following general principle.

###### Area Between Two Curves Using Vertical Slices

If two curves $y = g(x)$ and $y = f(x)$ intersect at $(a,g(a))$ and $(b,g(b))\text{,}$ and $g(x) \ge f(x)$ on the interval $[a,b]\text{,}$ then the area between the curves is

\begin{equation*} A = \int_a^b (g(x) - f(x)) \, dx\text{.} \end{equation*}
###### Example6.6

In each of the following problems, our goal is to determine the area of the region described. For each region, (i) determine the intersection points of the curves, (ii) sketch the region whose area is being found, (iii) draw and label a representative slice, and (iv) state the area of the representative slice. Then, state a definite integral whose value is the exact area of the region, and evaluate the integral to find the numeric value of the region's area.

1. The finite region bounded by $y = \sqrt{x}$ and $y = \frac{1}{4}x\text{.}$

2. The finite region bounded by $y = 12-2x^2$ and $y = x^2 - 8\text{.}$

3. The area bounded by the $y$-axis, $f(x) = \cos(x)\text{,}$ and $g(x) = \sin(x)\text{,}$ where we consider the region formed by the first positive value of $x$ for which $f$ and $g$ intersect.

4. The finite regions between the curves $y = x^3-x$ and $y = x^2\text{.}$

Hint
1. Find where the two curves intersect.

2. Plot both functions and think about how they compare to $y=x^2\text{.}$

3. Remember that $\sin(\pi/4)=\cos(\pi/4)\text{.}$

4. Find three points where the curves intersect and thus two finite regions between two pairs of those points.

1. $A = \int_{0}^{16} (\sqrt{x} - \frac{1}{4}x) \, dx = \frac{32}{3}\text{.}$

2. $A = \int_{-\sqrt{20}/3}^{\sqrt{20}/3} ((12-x^2)-(x^2-8)) \, dx \frac{160 \sqrt{\frac{5}{3}}}{3} \approx 68.853\text{.}$

3. $A = \int_0^\frac{\pi}{4} \left( \cos(x) - \sin(x) \right) dx = \sqrt{2} - 1\text{.}$

4. The left-hand region has area

\begin{equation*} A_1 = \int_{\frac{1 - \sqrt{5}}{2}}^0 \left( x^2 - \left(x^3 - x \right) \right) dx = \dfrac{13 + 5\sqrt{5}}{24} \approx 1.007514\text{.} \end{equation*}

The right-hand region has area

\begin{equation*} A_2 = \int_0^{\frac{1 + \sqrt{5}}{2}} \left( \left(x^3 - x \right) - x^2\right) dx = \dfrac{13 - 5\sqrt{5}}{24} \approx 0.075819\text{.} \end{equation*}
Solution
1. By solving $\sqrt{x}=\frac{1}{4}x\text{,}$ we see that the two curves intersect when $x=0$ and $x=16\text{,}$ so at the points $(0,0)$ and $(16,4)\text{.}$ We can plot the region as shown in the figure below. Slicing the region with thin vertical rectangles, we see that a typical rectangle's height is given by $\sqrt{x}-\frac{1}{4}x\text{,}$ and thus the area of such a slice is $A_{\text{rect} } = (\sqrt{x} - \frac{1}{4}x) \Delta x\text{.}$ It follows that the area of the region bounded by the two curves is

\begin{equation*} A = \int_{0}^{16} (\sqrt{x} - \frac{1}{4}x) \, dx\text{.} \end{equation*}

Evaluating the integral, we find $A = \frac{32}{3}\text{.}$

2. Solving $12-2x^2=x^2-8\text{,}$ we find that the two curves intersect where $3x^2=2-\text{,}$ so at $x=\pm \sqrt{20/3}\text{.}$ We can plot the region as shown in the figure below. Slicing the region with thin vertical rectangles, a typical rectangle's height is given by $(12-x^2)-(x^2-8)\text{,}$ and thus the area of such a slice is $A_{\text{rect} } = ((12-x^2)-(x^2-8)) \Delta x\text{.}$ Hence, the area of the region bounded by the two curves is

\begin{equation*} A = \int_{-\sqrt{20}/3}^{\sqrt{20}/3} ((12-x^2)-(x^2-8)) \, dx\text{.} \end{equation*}

Evaluating the integral, we find $A = \frac{160 \sqrt{\frac{5}{3}}}{3} \approx 68.853\text{.}$

3. The first positive value at which the sine and cosine functions intersect is $x=\frac{\pi}{4}\text{.}$ Plotting the functions in a surrounding region, we see the following figure. Slicing the region with vertical rectangles, we find the area of the shaded region is given by

\begin{equation*} A = \int_0^\frac{\pi}{4} \left( \cos(x) - \sin(x) \right) dx = \sqrt{2} - 1\text{.} \end{equation*}
4. The curves $y = x^3-x$ and $y = x^2$ intersect at three points, forming two regions. Setting $x^3-x=x^2$ and solving, we find that

\begin{equation*} x^3-x^2-x=x(x^2-x-1)=0\text{,} \end{equation*}

which implies that $x=0$ or $x = \frac{1}{2}(1 \pm \sqrt{5})\text{.}$ Plotting the curves, we see that for the left-hand region, the relative interval is $[\frac{1-\sqrt{5}}{2},0]$ and $g(x)=x^3-x \gt x^2 = f(x)\text{.}$ Thus, the left-hand region has area

\begin{equation*} A_1 = \int_{\frac{1 - \sqrt{5}}{2}}^0 \left( x^2 - \left(x^3 - x \right) \right) dx = \dfrac{13 + 5\sqrt{5}}{24} \approx 1.007514\text{.} \end{equation*}

For the right-hand region, the interval is $[0, \frac{1+\sqrt{5}}{2}]$ and $f(x)=x^2 \gt x^3-x = g(x)\text{,}$ so that its area is

\begin{equation*} A_2 = \int_0^{\frac{1 + \sqrt{5}}{2}} \left( \left(x^3 - x \right) - x^2\right) dx = \dfrac{13 - 5\sqrt{5}}{24} \approx 0.075819\text{.} \end{equation*}

### SubsectionFinding Area with Horizontal Slices

At times, the shape of a region may dictate that we use horizontal rectangular slices, instead of vertical ones.

###### Example6.7

Find the area of the region bounded by the parabola $x = y^2 - 1$ and the line $y = x-1\text{,}$ shown at left in Figure6.8. Figure6.8The area bounded by the functions $x = y^2-1$ and $y = x-1$ (at left), with the region sliced vertically (center) and horizontally (at right).
Solution

By solving the second equation for $x$ and writing $x = y + 1\text{,}$ we find that $y+1 = y^2 - 1\text{.}$ Hence the curves intersect where $y^2 - y - 2 = 0\text{.}$ Thus, we find $y = -1$ or $y = 2\text{,}$ so the intersection points of the two curves are $(0,-1)$ and $(3,2)\text{.}$

If we attempt to use vertical rectangles to slice up the area (as in the center graph of Figure6.8), we see that from $x = -1$ to $x = 0$ the curves that bound the top and bottom of the rectangle are one and the same, which could make the integral difficult. This suggests, as shown in the rightmost graph in the figure, that we try using horizontal rectangles.

Note that the width (horizontal distance) of a horizontal rectangle depends on $y\text{.}$ Between $y = -1$ and $y = 2\text{,}$ the right end of a representative rectangle is determined by the line $x = y+1\text{,}$ and the left end is determined by the parabola, $x = y^2-1\text{.}$ The thickness (vertical distance) of the rectangle is $\Delta y\text{.}$

Therefore, the area of the rectangle is

\begin{equation*} A_{\text{rect} } = [(y+1) - (y^2-1)] \Delta y\text{,} \end{equation*}

and the area between the two curves on the $y$-interval $[-1,2]$ is approximated by the Riemann sum

\begin{equation*} A \approx \sum_{i=1}^{n} [(y_i+1)-(y_i^2-1)] \Delta y\text{.} \end{equation*}

Taking the limit of the Riemann sum, it follows that the area of the region is

\begin{equation} A = \int_{y=-1}^{y=2} [(y+1) - (y^2-1)] \, dy\text{.}\label{E-IntWRTy}\tag{6.3} \end{equation}

We emphasize that we are integrating with respect to $y\text{;}$ this is because we chose to use horizontal rectangles whose widths depend on $y$ and whose thickness is denoted $\Delta y\text{.}$ It is a straightforward exercise to evaluate the integral in Equation(6.3) and find that $A = \frac{9}{2}\text{.}$

Just as with the use of vertical rectangles of thickness $\Delta x\text{,}$ we have a general principle for finding the area between two curves, which we state as follows.

###### Area Between Two Curves Using Horizontal Slices

If two curves $x = g(y)$ and $x = f(y)$ intersect at $(g(c),c)$ and $(g(d),d)\text{,}$ and $g(y) \ge f(y)$ on the interval $[c,d]\text{,}$ then the area between the curves is

\begin{equation*} A = \int_{y=c}^{y=d} (g(y) - f(y)) \, dy\text{.} \end{equation*}
###### Example6.9

In each of the following problems, our goal is to determine the area of the region described. For each region, (i) determine the intersection points of the curves, (ii) sketch the region whose area is being found, (iii) draw and label a representative slice, and (iv) state the area of the representative slice. Then, state a definite integral whose value is the exact area of the region, and evaluate the integral to find the numeric value of the region's area. Note well: At the step where you draw a representative slice, you need to make a choice about whether to slice vertically or horizontally.

1. The finite region bounded by $x=y^2$ and $x=6-2y^2\text{.}$

2. The finite region bounded by $x=1-y^2$ and $x = 2-2y^2\text{.}$

3. The area bounded by the $x$-axis, $y=x^2\text{,}$ and $y=2-x\text{.}$

4. The finite regions between the curves $x=y^2-2y$ and $y=x\text{.}$

Hint
1. Find where the curves intersect and plot them.

2. Observe that horizontal slices are appropriate, so we will integrate with respect to $y \text{.}$

3. Note that the region is approximately triangular with the $x$-axis as the base, $y=x^2$ as the left "side", and $y = 2-x$ as the right side.

4. Plot the region and observe that horizontal slices are appropriate because of the shape of the region.

1. $A = \int_{y=-\sqrt{2}}^{y=\sqrt{2}} (6-2y^2 - y^2) \, dy = 8\sqrt{2} \approx 11.314\text{.}$

2. $A = \int_{y=-1}^{y=1} (2-2y^2-(1-y^2)) \, dy = \frac{4}{3}\text{.}$

3. $A=\int_{y=0}^{y=1} \left((2-y) - \sqrt{y} \right) \, dy = \frac 56$

4. $A = \int_{0}^{3} (y - (y^2 - 2y)) \, dy = \frac{9}{2}\text{.}$

Solution
1. The two curves intersect when $y^2=6-2y^2\text{,}$ which is when $3y^2=6\text{.}$ Solving, they intersect at $y=\pm \sqrt{2}\text{.}$ Plotting the region, we see that for $y=-\sqrt{2}$ to $y=\sqrt{2}\text{,}$ $g(y) = 6-2y^2 \gt y^2 = f(y)\text{,}$ and so horizontal slices are appropriate for setting up the integral that represents the region's area. The area is thus given by

\begin{equation*} A = \int_{y=-\sqrt{2}}^{y=\sqrt{2}} (6-2y^2 - y^2) \, dy = 8\sqrt{2} \approx 11.314\text{.} \end{equation*}
2. The two curves intersect when $1-y^2=2-2y^2\text{,}$ which is when $3y^2=3\text{.}$ Solving, they intersect at $y=\pm 1\text{.}$ Graphing the region, on $-1 \lt y \lt 1\text{,}$ we see that $g(y) = 2-2y^2 \ge 1-y^2 = f(y)\text{.}$ Using horizontal slices, it follows that the region's area is given by

\begin{equation*} A = \int_{y=-1}^{y=1} (2-2y^2-(1-y^2)) \, dy = \int_{y=-1}^{y=1} (1-y^2) \, dy = \frac{4}{3}\text{.} \end{equation*}
3. Standard algebra shows that the two curves meet at $(1,1)\text{.}$ A careful plot of the region shows that the two curves, together with the $x$-axis, form a roughly triangular region with the $x$-axis as the base, $y=x^2$ as the left side, and $y = 2-x$ as the right side. Using horizontal slices, and writing each respective equation with $x$ as a function of $y$ (so $x=\sqrt{y}$ and $x=2-y$), we find that the area of this region is

\begin{equation*} A=\int_{y=0}^{y=1} \left((2-y) - \sqrt{y} \right) \, dy = \\ \left. 2y- \frac12 y^2- \frac 23 y^{3/2} \right|_0^1=\frac 56 \end{equation*}
4. The two curves intersect at $(0,0)$ and $(3,3)\text{.}$ For $0 \le y \le 3\text{,}$ we observe that $g(y)=y \ge y^2-2y = f(y)\text{,}$ so using horizontal slices and the corresponding definite integral, we find that the region's area is

\begin{equation*} A = \int_{0}^{3} (y - (y^2 - 2y)) \, dy = \int_{0}^{3} (3y - y^2) \, dy = \frac{9}{2}\text{.} \end{equation*}

### SubsectionFinding the Length of a Curve

We can also use the definite integral to find the length of a portion of a curve. We use the same fundamental principle: we slice the curve up into small pieces whose lengths we can easily approximate. Specifically, we subdivide the curve into small approximating line segments, as shown at left in Figure6.10. Figure6.10At left, a continuous function $y = f(x)$ whose length we seek on the interval $a = x_0$ to $b = x_3\text{.}$ At right, a close up view of a portion of the curve.

We estimate the length $L_{\text{slice} }$ of each portion of the curve on a small interval of length $\Delta x\text{.}$ We use the right triangle with legs parallel to the coordinate axes and hypotenuse connecting the endpoints of the slice, as seen at right in Figure6.10. The length, $h\text{,}$ of the hypotenuse approximates the length, $L_{\text{slice} }\text{,}$ of the curve between the two selected points. Thus,

\begin{equation*} L_{\text{slice} } \approx h = \sqrt{ (\Delta x)^2 + (\Delta y)^2 }\text{.} \end{equation*}

Next we use algebra to rearrange the expression for the length of the hypotenuse into a form that we can integrate. By removing a factor of $(\Delta x)^2\text{,}$ we find

\begin{align*} L_{\text{slice}} &\approx \sqrt{ (\Delta x)^2 + (\Delta y)^2 }\\ &= \sqrt{ (\Delta x)^2\left(1 + \frac{(\Delta y)^2}{(\Delta x)^2} \right)}\\ &= \sqrt{1 + \frac{(\Delta y)^2}{(\Delta x)^2} } \cdot \Delta x\text{.} \end{align*}

Then, as $n \to \infty$ and $\Delta x \to 0\text{,}$ we have that $\frac{\Delta y}{\Delta x} \to \frac{dy}{dx} = f'(x)\text{.}$ Thus, we can say that

\begin{equation*} L_{\text{slice} } \approx \sqrt{1 + f'(x)^2} \Delta x\text{.} \end{equation*}

Taking a Riemann sum of all of these slices and letting $n \to \infty\text{,}$ we arrive at the following fact.

###### Arc Length

Given a differentiable function $f$ on an interval $[a,b]\text{,}$ the total arc length, $L\text{,}$ along the curve $y = f(x)$ from $x = a$ to $x = b$ is given by

\begin{equation*} L = \int_a^b \sqrt{1+f'(x)^2} \, dx\text{.} \end{equation*}
###### Example6.11

Each of the following questions somehow involves the arc length along a curve.

1. Use the definition and appropriate computational technology to determine the arc length along $y = x^2$ from $x = -1$ to $x = 1\text{.}$

2. Find the arc length of $y = \sqrt{4-x^2}$ on the interval $-2 \le x \le 2\text{.}$ Find this value in two different ways: (a) by using a definite integral, and (b) by using a familiar property of the curve.

3. Determine the arc length of $y = xe^{3x}$ on the interval $[0,1]\text{.}$

4. Will the integrals that arise calculating arc length typically be ones that we can evaluate exactly using the First FTC, or ones that we need to approximate? Why?

5. A moving particle is traveling along the curve given by $y = f(x) = 0.1x^2 + 1\text{,}$ and does so at a constant rate of 7 cm/sec, where both $x$ and $y$ are measured in cm (that is, the curve $y = f(x)$ is the path along which the object actually travels; the curve is not a position function). Find the position of the particle when $t = 4$ sec, assuming that when $t = 0\text{,}$ the particle's location is $(0,f(0))\text{.}$

Hint
1. Recall that $L = \int_a^b \sqrt{1+f'(x)^2} \, dx\text{.}$

2. Remember that $x^2 + y^2 = 4$ generates a circle centered at $(0,0)$ of radius 4.

3. Apply the arc length formula. Use technology to evaluate the integral.

4. Most expressions involving square roots are difficult to antidifferentiate.

5. Here you can determine the arc length and then experiment to find the upper limit of integration, which will help you determine position.

1. $L \approx 2.95789\text{.}$

2. $L = \int_{-2}^{2} \sqrt{\frac{4}{4-x^2}} \, dx = 2\pi\text{.}$

3. $L = \int_0^1 \sqrt{1 + e^{6x}(9x^2 + 6x + 1)} \, dx \approx 20.1773\text{.}$

4. We will usually have to estimate the value of $\int_a^b \sqrt{1+f'(x)^2} \, dx$ using computational technology.

5. Approximately $(9.011,f(9.011)) = (9.011, 9.1198)\text{.}$

Solution
1. Using the formula for arc length, we know

\begin{equation*} L = \int_{-1}^{1} \sqrt{1+(2x)^2} \,dx\text{.} \end{equation*}

Evaluating the integral, we find $L \approx 2.95789\text{.}$

2. The curve is a semi-circle of radius 2, so its length must be $L = \frac{1}{2} \cdot 2 \pi r = 2\pi\text{.}$ This is confirmed by the arc length formula, since $f(x) = \sqrt{4-x^2}\text{,}$ so $f'(x) = \frac{1}{2}(4-x^2)^{-1/2}(-2x)$

\begin{equation*} L = \int_{-2}^{2} \sqrt{1 + \left(-x(4-x^2)^{-1/2} \right)^2} \, dx = \int_{-2}^{2} \sqrt{1 + x^2(4-x^2)^{-1}} \, dx\text{.} \end{equation*}

It follows1This integral is actually improper because the integrand is undefined at the endpoints, $x = \pm 2\text{.}$ We learned how to evaluate such integrals in Section5.10. that

\begin{equation*} L = \int_{-2}^{2} \sqrt{\frac{4}{4-x^2}} \, dx = 2\pi\text{.} \end{equation*}
3. For $y = xe^{3x}\text{,}$ we observe that $y' = 3xe^{3x} + e^{3x} = e^{3x}(3x+1)\text{,}$ and that

\begin{equation*} (y')^2 = e^{6x}(9x^2 + 6x + 1)\text{.} \end{equation*}

Thus, the arc length of the curve on $[0,1]$ is

\begin{equation*} L = \int_0^1 \sqrt{1 + e^{6x}(9x^2 + 6x + 1)} \, dx \approx 20.1773\text{.} \end{equation*}
4. Because $\sqrt{1+f'(x)^2}$ will rarely simplify nicely and rarely have an elementary antiderivative, we will usually have to estimate the value of $\int_a^b \sqrt{1+f'(x)^2} \, dx$ using computational technology.

5. After 4 seconds traveling at 7 cm/sec, the particle has moved 28 cm. So, we have to find the value of $b$ for which the arc length $L$ along $[0,b]$ is 28. Thus, we know that

\begin{equation*} 28 = L = \int_0^b \sqrt{1 + (0.2x)^2} \, dx\text{.} \end{equation*}

Rewriting the integral, we see

\begin{equation*} 28 = \int_0^b \sqrt{1+0.4x^2} \, dx\text{.} \end{equation*}

Experimenting with different values of $b$ in a computational engine like Wolfram|Alpha, we find that for $b \approx 9.011\text{,}$ $\int_0^{9.011} \sqrt{1+0.4x^2} \, dx \approx 27.9992\text{,}$ and thus the position of the particle when $t=4$ is approximately $(9.011,f(9.011)) = (9.011, 9.1198)\text{.}$

The same technique can be applied to parametric curves. For now consider a curve $(x(t), y(t))$ travelling in just two dimensions (but see Example6.14 for curves in 3D). As before, the length of a small segment will be $\sqrt{(\Delta x)^2+(\Delta y)^2}\text{,}$ but here we will factor out $\Delta t$ to get:

\begin{equation*} L_{slice}\approx \sqrt{\left(\frac{\Delta x}{\Delta t} \right)^2+\left(\frac{\Delta y}{\Delta t} \right)^2} \cdot \Delta t \end{equation*}

Now, taking the limit as $\Delta t \rightarrow 0 \text{,}$ $\frac{\Delta x}{\Delta t}$ and $\frac{\Delta y}{\Delta t}$ turn into $\frac{dx}{dt}$ and $\frac{dy}{dt}\text{,}$ so the length of the segment is

\begin{equation*} L=\int_{t=a}^{t=b} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\, dt \text{.} \end{equation*}

We summarize our result here:

###### Arc Length for a Parametric Curve

Given two differentiable functions $x(t), y(t)$ on an interval $a \leq t \leq b\text{,}$ the total arc length $L$ along the parametric curve $(x(t), y(t))$ from $t = a$ to $t = b$ is given by

\begin{equation*} L=\int_{t=a}^{t=b} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} dt\text{.} \end{equation*}
###### Example6.12
1. Recall that the unit circle is traced out by the parametric equations $x(t)=\cos(t)$ and $y(t)=\sin(t)$ on the interval $0 \leq t \leq 2 \pi\text{.}$ Set up and evaluate the arc length integral to find the circumference of the unit circle.

2. An ellipse (a stretched circle) can be traced out by the parametric equations $x(t)=2\cos(t)$ and $y(t)=\sin(t)\text{.}$ Set up the arc length integral to find the perimeter of this ellipse. You do not have to evaluate the integral.

Hint
1. Use the formula for the arc length for a parametric curve to set up the integral. Recall also the Pythagorean Identity for trig functions: $\sin^2(t)+\cos^2(t)=1\text{.}$

2. Use the formula for the arc length for a parametric curve to set up the integral.

1. $L= \displaystyle \int_0^{2\pi}1 \, dt=2\pi$

2. $L= \int_0^{2\pi}\sqrt{(-2\sin(t)^2)+(\cos(t))^2} \, dt$

Solution
1. Since $\frac{dx}{dt}= -\sin(t)$ and $\frac{dy}{dt}= \cos(t)\text{,}$ then

\begin{equation*} L= \int_0^{2\pi}\sqrt{(-\sin(t)^2)+(\cos(t))^2} \, dt = \int_0^{2\pi}1 \, dt=2\pi \end{equation*}
2. Since $\frac{dx}{dt}= -2\sin(t)$ and $\frac{dy}{dt}= \cos(t)\text{,}$ then

\begin{equation*} L= \int_0^{2\pi}\sqrt{(-2\sin(t)^2)+(\cos(t))^2} \, dt = \int_0^{2\pi}\sqrt{4\sin^2(t)+\cos^2(t)} \, dt \end{equation*}

We do not need to evaluate this integral, and in fact it is an example of an elliptical integral of the second kind, which are a famous class of integrals without elementary antiderivatives.

###### Example6.13

Alex drives to work, and their route is along a path given by $x(t)=t^3\text{,}$ $y(t)=t^4+t$ for $0 \leq t \leq 1\text{.}$ How long is their route to work? (You may wish to use a calculator or computer to numerically evaluate the integral.)

Hint

Set up the integral using formula for arc length for a parametric curve. Then use a computer to evaluate the integral.

$L \approx 2.266$

Solution

Since $\frac{dx}{dt}= 3t^2$ and $\frac{dy}{dt}= 4t^3+1\text{,}$ then

\begin{equation*} L= \int_0^{1}\sqrt{(3t^2)^2+(4t^3+1)^2} \, dt \approx 2.266 \end{equation*}
###### Example6.14

Suppose that one has a parametric curve travelling through 3 dimensions given by $f(t)=(x(t), y(t), z(t))\text{.}$ How would you change the Arc Length for a Parametric Curve formula to include the third dimension?

Hint

Treat the $z$-coordinate the same way the $x$- and $y$-coordinates are treated.

$\displaystyle L=\int_{t=a}^{t=b} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2} dt$
Solution

The length of a 3D line segment is given by $\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2}\text{,}$ so factoring out a factor of $\Delta t$ and taking the limit as $\Delta t \rightarrow 0\text{,}$ we get

\begin{equation*} L=\int_{t=a}^{t=b} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2} dt \end{equation*}

### SubsectionSummary

• To find the area between two curves, we think about slicing the region into thin rectangles. If, for instance, the area of a typical rectangle on the interval $x = a$ to $x = b$ is given by $A_{\text{rect} } = (g(x) - f(x)) \Delta x\text{,}$ then the exact area of the region is given by the definite integral

\begin{equation*} A = \int_a^b (g(x)-f(x))\, dx\text{.} \end{equation*}
• The shape of the region usually dictates whether we should use vertical rectangles of thickness $\Delta x$ or horizontal rectangles of thickness $\Delta y\text{.}$ We want the height of the rectangle given by the difference between two curves: if those curves are best thought of as functions of $y\text{,}$ we use horizontal rectangles, whereas if those curves are best viewed as functions of $x\text{,}$ we use vertical rectangles.

• The arc length, $L\text{,}$ along the curve $y = f(x)$ from $x = a$ to $x = b$ is given by

\begin{equation*} L = \int_a^b \sqrt{1 + f'(x)^2} \, dx\text{.} \end{equation*}

The definite integral may also be used to find arc length of parametric curves.

### SubsectionExercises

Find the exact area of each described region.

1. The finite region between the curves $x = y(y-2)$ and $x=-(y-1)(y-3)\text{.}$

2. The region between the sine and cosine functions on the interval $[\frac{\pi}{4}, \frac{3\pi}{4}]\text{.}$

3. The finite region between $x = y^2 - y - 2$ and $y = 2x-1\text{.}$

4. The finite region between $y = mx$ and $y = x^2-1\text{,}$ where $m$ is a positive constant.

Let $f(x) = 1-x^2$ and $g(x) = ax^2 - a\text{,}$ where $a$ is an unknown positive real number. For what value(s) of $a$ is the area between the curves $f$ and $g$ equal to 2?

Let $f(x) = 2-x^2\text{.}$ Recall that the average value of any continuous function $f$ on an interval $[a,b]$ is given by $\frac{1}{b-a} \int_a^b f(x) \, dx\text{.}$

1. Find the average value of $f(x) = 2-x^2$ on the interval $[0,\sqrt{2}]\text{.}$ Call this value $r\text{.}$

2. Sketch a graph of $y = f(x)$ and $y = r\text{.}$ Find their intersection point(s).

3. Show that on the interval $[0,\sqrt{2}]\text{,}$ the amount of area that lies below $y = f(x)$ and above $y = r$ is equal to the amount of area that lies below $y = r$ and above $y = f(x)\text{.}$

4. Will the result of (c) be true for any continuous function and its average value on any interval? Why?

The parametric function given by $x(t)= \sin(2t)\text{,}$ $y(t)= \sin(t)$ traces out a figure-8 pattern for $0 \leq t \leq 2\pi\text{.}$ Find the length of this curve. (You may wish to use a calculator or computer to numerically evaluate the integral.)

Hint

Set up the integral using the Arc Length for a Parametric Curve formula, then use a computer to evaluate the integral.