For $f(x)=x^{2}\left(8+x^{3}\right)^{10}$ , find a function $F(x)$ such that $F^{\prime}(x)=f(x)$ and $F(0)=0$ .
A) $\frac{1}{33}\left(\left(8+x^{3}\right)^{11}\right)$
B) $\frac{1}{33}\left(\left(8+x^{3}\right)^{11}-8^{11}\right)$
C) $\frac{1}{99}\left(\left(8+x^{3}\right)^{11}-8^{11}\right)$
D) $\frac{x}{99}\left(8+x^{3}\right)^{11}$

Question

Quizplus · Accepted Answer

To find a function $F(x)$ such that $F'(x) = f(x)$, we integrate $f(x) = x^2(8+x^3)^{10}$. Using the substitution method, let $u = 8+x^3$, then $du = 3x^2 dx$, and $x^2 dx = \frac{1}{3}du$. Integrating, we get $F(x) = \frac{1}{3} \cdot \frac{(8+x^3)^{11}}{11} + C$, simplifying to $F(x) = \frac{1}{33}(8+x^3)^{11} + C$. To find $C$, we use $F(0) = 0$, which gives $C = -\frac{8^{11}}{33}$. Thus, $F(x) = \frac{1}{33}((8+x^3)^{11} - 8^{11})$.

For f(x)=x2(8+x3)10f(x)=x^{2}\left(8+x^{3}\right)^{10}f(x)=x2(8+x3)10 , Find a Function F(x)F(x)F(x) Such That F′(x)=f(x)F^{\prime}(x)=f(x)F′(x)=f(x) And F(0)=0F(0)=0F(0)=0 A) 133((8+x3)11)\frac{1}{33}\left(\left(8+x^{3}\right)^{11}\right)331​((8+x3)11)

For $f(x)=x^{2}\left(8+x^{3}\right)^{10}$ , Find a Function $F(x)$ Such That $F^{\prime}(x)=f(x)$ And $F(0)=0$
A) $\frac{1}{33}\left(\left(8+x^{3}\right)^{11}\right)$