Coordinated Calculus

1. If you don't recall it, review the statement of the Second FTC above.

2. Note that \(A(1)= \int_2^1 f(t) \, dt\text{.}\)

3. Don't miss our key conclusion from (a).

4. Compare the values of \(A(1)\) and \(F(1)\text{.}\)

5. What does the Second FTC tell us about the relationship between \(B\) and \(f\text{?}\)

1. \(A'(x) = f(x)\text{.}\)

2. \(A(1) = -\frac{\pi}{4}\text{.}\)

3. \(A\) is increasing wherever \(f\) is positive; \(A\) is CCU wherever \(f\) is increasing. \(A(2) = 0\text{,}\) \(A(3) = -0.5\text{,}\) \(A(4) = -1.5\text{,}\) \(A(5) = -2\text{,}\) \(A(6) = -2 + \frac{\pi}{4}\text{,}\) and \(A(7) = -2 + \frac{\pi}{2}\text{.}\)

4. \(F\) and \(A\) differ by the constant \(\frac{\pi}{4} - \frac{1}{2}\text{.}\)

5. \(B\) and \(C\) have the same shape as \(A\) and \(F\text{,}\) and differ from \(A\) by a constant. Observe that \(B(3) = 0\) and \(C(1) = 0\text{.}\)

1. By the Second FTC, \(A'(x) = f(x)\text{.}\)

2. Since \(A(1)= \int_2^1 f(t) \, dt = -\int_1^2 f(t) \, dt\text{,}\) it follows \(A(1) = -\frac{\pi}{4}\text{.}\)

3. Note that \(A\) is increasing wherever \(f\) is positive, and \(A\) is CCU wherever \(f\) is increasing. Similar conclusions follow for \(A\) being decreasing and/or concave down. Moreover, \(A(2) = 0\text{,}\) \(A(3) = -0.5\text{,}\) \(A(4) = -1.5\text{,}\) \(A(5) = -2\text{,}\) \(A(6) = -2 + \frac{\pi}{4}\text{,}\) and \(A(7) = -2 + \frac{\pi}{2}\text{.}\)

4. In our current example, \(A\) is an antiderivative of \(f\) that satisfies \(A(0) = -\frac{1}{2} - \frac{\pi}{4}\text{.}\) Our earlier work with \(F\) showed that \(F\) is an antiderivative of \(F\) that satisfied \(F(0) = -1\text{.}\) Since \(F\) and \(A\) are both antiderivatives of \(f\text{,}\) they differ by a constant, and that constant is \(-1 - (-\frac{1}{2} - \frac{\pi}{4}) = \frac{\pi}{4} - \frac{1}{2}\text{.}\)

5. The Second FTC tells us that \(B' = f\) and \(C' = f\text{.}\) Thus, \(B\) and \(C\) are each antiderivatives of \(f\text{,}\) have the same shape as \(A\) and \(F\text{,}\) and each differ from \(A\) by just a constant. Observing that \(B(3) = 0\) and \(C(1) = 0\) enables us to easily sketch these shifted versions of \(A\text{.}\)

\(E\) is closely related to the well known error function ³The error function is defined by the rule \(\erf (x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} \,dt\) and has the key property that \(0 \le \erf (x) \lt 1\) for all \(x \ge 0\) and moreover that \(\lim_{x \to \infty} \erf (x) = 1\text{.}\) in probability and statistics. It turns out that the function \(e^{-t^2}\) does not have an elementary antiderivative.

While we cannot evaluate \(E\) exactly for any value other than \(x = 0\text{,}\) we still can gain a tremendous amount of information about the function \(E\text{.}\) By applying the rule in Equation(5.4) to \(E\text{,}\) it follows that

so we know a formula for the derivative of \(E\text{,}\) and we know that \(E(0) = 0\text{.}\) This information is precisely the type we were given in Example3.7, where we were given information about the derivative of a function, but lacked a formula for the function itself.

Using the first and second derivatives of \(E\text{,}\) along with the fact that \(E(0) = 0\text{,}\) we can determine more information about the behavior of \(E\text{.}\) First, we note that for all real numbers \(x\text{,}\) \(e^{-x^2} \gt 0\text{,}\) and thus \(E'(x) \gt 0\) for all \(x\text{.}\) Thus \(E\) is an always increasing function. Further, as \(x \to \infty\text{,}\) \(E'(x) = e^{-x^2} \to 0\text{,}\) so the slope of the function \(E\) tends to zero as \(x \to \infty\) (and similarly as \(x \to -\infty\)). Indeed, it turns out that \(E\) has horizontal asymptotes as \(x\) increases or decreases without bound.

In addition, we can observe that \(E''(x) = -2xe^{-x^2}\text{,}\) and that \(E''(0) = 0\text{,}\) while \(E''(x) \lt 0\) for \(x \gt 0\) and \(E''(x) \gt 0\) for \(x \lt 0\text{.}\) This information tells us that \(E\) is concave up for \(x\lt 0\) and concave down for \(x \gt 0\) with a point of inflection at \(x = 0\text{.}\)

The only thing we lack at this point is a sense of how big \(E\) can get as \(x\) increases. If we use a midpoint Riemann sum with 10 subintervals to estimate \(E(2)\text{,}\) we see that \(E(2) \approx 0.8822\text{;}\) a similar calculation to estimate \(E(3)\) shows little change (\(E(3) \approx 0.8862\)), so it appears that as \(x\) increases without bound, \(E\) approaches a value just larger than \(0.886\text{,}\) which aligns with the fact that \(E\) has horizontal asymptotes. Putting all of this information together (and using the symmetry of \(f(t) = e^{-t^2}\)), we see the results shown in Figure5.32.

Because \(E\) is the antiderivative of \(f(t) = e^{-t^2}\) that satisfies \(E(0) = 0\text{,}\) values on the graph of \(y = E(x)\) represent the net signed area of the region bounded by \(f(t) = e^{-t^2}\) from 0 up to \(x\text{.}\) We see that the value of \(E\) increases rapidly near zero but then levels off as \(x\) increases, since there is less and less additional accumulated area bounded by \(f(t) = e^{-t^2}\) as \(x\) increases.

1. Use computing technology appropriately to generate the desired graph.

2. Again, recall the statement of the Second FTC.

3. Where is \(F'\) positive? \(F'\) negative?

4. Note that \(F'' = f'\text{.}\)

5. Remember that \(F(5) = \int_0^5 \frac{t}{1+t^2} \, dt\text{.}\)

6. Don't forget that \(F(0) = 0\text{.}\)

1. See the plot at below left.

2. \(F' = f\text{.}\)

3. \(F\) is increasing for all \(x \gt 0\text{;}\) \(F\) is decreasing for \(x \lt 0\)

4. \(F\) is CCU on \(-1 \lt x \lt 1\) and CCD for \(x \lt -1\) and \(x \gt 1\text{.}\)

5. \(F(5) \approx 2.35973\text{;}\) \(F(10) \approx 2.35973\text{.}\)

6. See the graph shown below on the right.

1. See the plot at below left.

2. \(F' = f\text{,}\) by the Second FTC.

3. \(F\) is increasing wherever \(F'=f\) is positive, so for all \(x \gt 0\text{.}\) Similarly, \(F\) is decreasing for \(x \lt 0\)

4. \(F\) is concave up wherever \(F' = f\) is increasing or wherever \(F'' = f'\) is positive. It is straightforward to show that \(f''\) is positive for \(-1 \lt x \lt 1\) and negative otherwise, thus \(F\) is concave up on \(-1 \lt x \lt 1\) and concave down for \(x \lt -1\) and \(x \gt 1\text{.}\)

5. \(F(5) = \int_0^5 \frac{t}{1+t^2} \, dt \approx 2.35973\text{,}\) using a midpoint Riemann sum with 10 subintervals. Similarly, \(F(10) = \int_0^{10} \frac{t}{1+t^2} \, dt \approx 2.35973\text{.}\)

6. Recalling that \(F(0) = 0\) and using the values and information we've found in (b)-(e), we arrive at the graph at below right.

1. Second FTC.

2. First FTC.

3. \(\int_{x}^{1} g(t) \, dt = -\int_{1}^{x} g(t) \, dt\text{.}\)

4. First FTC.

5. Let \(F(x) = \int_4^x \sin(t^2) \, dt\) and observe that this problem is asking you to evaluate \(\frac{d}{dx} \left[F(x^3)] \right]\text{.}\)

1. \(\frac{d}{dx} \left[ \int_4^x e^{t^2} \, dt \right] = e^{x^2}\text{.}\)

2. \(\int_{-2}^x \frac{d}{dt} \left[ \frac{t^4}{1+t^4} \right] \, dt = \frac{x^4}{1+x^4} - \frac{8}{9}\text{.}\)

3. \(\frac{d}{dx} \left[ \int_{x}^1 \cos(t^3) \, dt \right] = -\cos(x^3)\text{.}\)

4. \(\int_{3}^x \frac{d}{dt} \left[ \ln(1+t^2) \right] \, dt = \ln(1+x^2)-\ln(4)\text{.}\)

5. \(\frac{d}{dx} \left[ \int_4^{x^3} \sin(t^2) \, dt \right] = \sin(x^6) \cdot 3x^2\text{.}\)

1. By the Second FTC, \(\frac{d}{dx} \left[ \int_4^x e^{t^2} \, dt \right] = e^{x^2}\text{.}\)

2. By the First FTC, \(\int_{-2}^x \frac{d}{dt} \left[ \frac{t^4}{1+t^4} \right] \, dt = \left. \left[ \frac{t^4}{1+t^4} \right] \right|_{-2}^{x} = \frac{x^4}{1+x^4} - \frac{16}{17}\text{.}\)

3. Since \(\int_{x}^{1} g(t) \, dt = -\int_{1}^{x} g(t) \, dt\text{,}\) it follows by this fact and the Second FTC that \(\frac{d}{dx} \left[ \int_{x}^1 \cos(t^3) \, dt \right] = -\frac{d}{dx} \left[ \int_{1}^x \cos(t^3) \, dt \right] = -\cos(x^3)\text{.}\)

4. By the First FTC, \(\int_{3}^x \frac{d}{dt} \left[ \ln(1+t^2) \right] \, dt = \left. \ln(1+t^2) \right|_{3}^{x} = \ln(1+x^2)-\ln(10)\text{.}\)

5. Letting \(F(x) = \int_4^x \sin(t^2) \, dt\) it follows that we need to compute \(\frac{d}{dx} \left[F(x^3)] \right]\text{.}\) By the Chain Rule, \(\frac{d}{dx} \left[F(x^3)] \right] = F'(x^3) \cdot 3x^2\text{.}\) By the Second FTC, we know that \(F'(x) = \sin(x^2)\text{,}\) and thus \(\frac{d}{dx} \left[ \int_4^{x^3} \sin(t^2) \, dt \right] = \frac{d}{dx} \left[F(x^3)] \right] = \sin(x^6) \cdot 3x^2\text{.}\)