Math  /  Calculus

Question7-18 Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
7. g(x)=1x1t3+1dtg(x)=\int_{1}^{x} \frac{1}{t^{3}+1} d t
8. g(x)=3xet2tdtg(x)=\int_{3}^{x} e^{t^{2}-t} d t
9. g(s)=5s(tt2)8dtg(s)=\int_{5}^{s}\left(t-t^{2}\right)^{8} d t
10. g(r)=0rx2+4dxg(r)=\int_{0}^{r} \sqrt{x^{2}+4} d x
11. F(x)=xπ1+sectdtF(x)=\int_{x}^{\pi} \sqrt{1+\sec t} d t [\left[\right. Hint: xπ1+sectdt=πx1+sectdt]\left.\int_{x}^{\pi} \sqrt{1+\sec t} d t=-\int_{\pi}^{x} \sqrt{1+\sec t} d t\right]

Studdy Solution
Apply Part 1 of the Fundamental Theorem of Calculus to find the derivative.
Now, let's solve each problem:
**Problem 7:**
STEP_1: The function inside the integral is 1t3+1 \frac{1}{t^3 + 1} .
STEP_2: By the Fundamental Theorem of Calculus, the derivative is:
g(x)=1x3+1 g'(x) = \frac{1}{x^3 + 1}
**Problem 8:**
STEP_1: The function inside the integral is et2t e^{t^2 - t} .
STEP_2: By the Fundamental Theorem of Calculus, the derivative is:
g(x)=ex2x g'(x) = e^{x^2 - x}
**Problem 9:**
STEP_1: The function inside the integral is (tt2)8 (t - t^2)^8 .
STEP_2: By the Fundamental Theorem of Calculus, the derivative is:
g(s)=(ss2)8 g'(s) = (s - s^2)^8
**Problem 10:**
STEP_1: The function inside the integral is x2+4 \sqrt{x^2 + 4} .
STEP_2: By the Fundamental Theorem of Calculus, the derivative is:
g(r)=r2+4 g'(r) = \sqrt{r^2 + 4}
**Problem 11:**
STEP_1: The function inside the integral is 1+sect \sqrt{1 + \sec t} .
STEP_2: Since the integral is from x x to π \pi , we use the hint to rewrite it as:
F(x)=πx1+sectdt F(x) = -\int_{\pi}^{x} \sqrt{1 + \sec t} \, dt
By the Fundamental Theorem of Calculus, the derivative is:
F(x)=1+secx F'(x) = -\sqrt{1 + \sec x}
The derivatives for each problem are:
7. g(x)=1x3+1 g'(x) = \frac{1}{x^3 + 1}
8. g(x)=ex2x g'(x) = e^{x^2 - x}
9. g(s)=(ss2)8 g'(s) = (s - s^2)^8
10. g(r)=r2+4 g'(r) = \sqrt{r^2 + 4}
11. F(x)=1+secx F'(x) = -\sqrt{1 + \sec x}

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