Activity 62.
Evaluate each of the following derivatives and definite integrals. Clearly cite whether you use the First or Second FTC in so doing.
(a)
\(\displaystyle \frac{d}{dx} \left[ \int_4^x e^{t^2} \, dt \right]\)
Hint.
Second FTC.
Answer.
\(\displaystyle \frac{d}{dx} \left[ \int_4^x e^{t^2} \, dt \right] = e^{x^2}\text{.}\)
Solution.
By the Second FTC, \(\frac{d}{dx} \left[ \int_4^x e^{t^2} \, dt \right] = e^{x^2}\text{.}\)
(b)
\(\displaystyle \int_{-2}^x \frac{d}{dt} \left[ \frac{t^4}{1+t^4} \right] \, dt\)
Hint.
First FTC.
Answer.
\(\displaystyle \int_{-2}^x \frac{d}{dt} \left[ \frac{t^4}{1+t^4} \right] \, dt = \frac{x^4}{1+x^4} - \frac{16}{17}\text{.}\)
Solution.
By the First FTC, \(\int_{-2}^x \frac{d}{dt} \left[ \frac{t^4}{1+t^4} \right] \, dt = \left. \left[ \frac{t^4}{1+t^4} \right] \right|_{-2}^{x} = \frac{x^4}{1+x^4} - \frac{16}{17}\text{.}\)
(c)
\(\displaystyle \frac{d}{dx} \left[ \int_{x}^1 \cos(t^3) \, dt \right]\)
Hint.
\(\displaystyle \int_{x}^{1} g(t) \, dt = -\int_{1}^{x} g(t) \, dt\text{.}\)
Answer.
\(\displaystyle \frac{d}{dx} \left[ \int_{x}^1 \cos(t^3) \, dt \right] = -\cos(x^3)\text{.}\)
Solution.
Since \(\int_{x}^{1} g(t) \, dt = -\int_{1}^{x} g(t) \, dt\text{,}\) it follows by this fact and the Second FTC that \(\frac{d}{dx} \left[ \int_{x}^1 \cos(t^3) \, dt \right] = -\frac{d}{dx} \left[ \int_{1}^x \cos(t^3) \, dt \right] = -\cos(x^3)\text{.}\)
(d)
\(\displaystyle \int_{3}^x \frac{d}{dt} \left[ \ln(1+t^2) \right] \, dt\)
Hint.
First FTC.
Answer.
\(\displaystyle \int_{3}^x \frac{d}{dt} \left[ \ln(1+t^2) \right] \, dt = \ln(1+x^2)-\ln(10)\text{.}\)
Solution.
By the First FTC, \(\int_{3}^x \frac{d}{dt} \left[ \ln(1+t^2) \right] \, dt = \left. \ln(1+t^2) \right|_{3}^{x} = \ln(1+x^2)-\ln(10)\text{.}\)
(e)
\(\displaystyle \frac{d}{dx} \left[ \int_4^{x^3} \sin(t^2) \, dt \right]\)
Hint.
Let \(F(x) = \int_4^x \sin(t^2) \, dt\) and observe that this problem is asking you to evaluate \(\frac{d}{dx} \left[F(x^3)] \right]\text{.}\)
Answer.
\(\displaystyle \frac{d}{dx} \left[ \int_4^{x^3} \sin(t^2) \, dt \right] = \sin(x^6) \cdot 3x^2\text{.}\)
Solution.
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 \(\frac{d}{dx} \left[ \int_4^{x^3} \sin(t^2) \, dt \right] = \frac{d}{dx} \left[F(x^3)] \right] = \sin(x^6) \cdot 3x^2\text{.}\)
