For $f(x)=x^{2} e^{10 x}$ , find a function $F(x)$ such that $F^{\prime}(x)=f(x)$ and $F(0)=0$ .

Question

Quizplus · Accepted Answer

To find a function $F(x)$ such that $F'(x) = f(x) = x^2e^{10x}$, we integrate $f(x)$ with respect to $x$. Using integration by parts multiple times or recognizing this as a product of a polynomial and an exponential function, we find that the antiderivative is $e^{10x}\left(\frac{1}{10}x^2 - \frac{1}{50}x + \frac{1}{500}\right) + C$. To find $C$, we use the condition $F(0) = 0$, which gives $C = -\frac{1}{500}$. Thus, the correct function is $e^{10x}\left(\frac{1}{10}x^2 - \frac{1}{50}x + \frac{1}{500}\right) - \frac{1}{500}$.

For f(x)=x2e10xf(x)=x^{2} e^{10 x}f(x)=x2e10x , 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

For $f(x)=x^{2} e^{10 x}$ , Find a Function $F(x)$ Such That $F^{\prime}(x)=f(x)$ And $F(0)=0$