If int x^2 cdot e^x dx = e^x f(x) + c, then t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Step $1$: Evaluate the integral $\int x^2 e^x dx$ using integration by parts $\int u dv = uv - \int v du$.Let $u = x^2$ and $dv = e^x dx$. Then $d

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Step $1$: Evaluate the integral $\int x^2 e^x dx$ using integration by parts $\int u dv = uv - \int v du$.Let $u = x^2$ and $dv = e^x dx$. Then $du = 2x dx$ and $v = e^x$.$\int x^2 e^x dx = x^2 e^x - \int 2x e^x dx$.Step $2$: Evaluate $\int 2x e^x dx$ using integration by parts again.Let $u = 2x$ and $dv = e^x dx$. Then $du = 2 dx$ and $v = e^x$.$\int 2x e^x dx = 2x e^x - \int 2 e^x dx = 2x e^x - 2e^x$.Step $3$: Substitute back into the original integral.$\int x^2 e^x dx = x^2 e^x - (2x e^x - 2e^x) + c = e^x(x^2 - 2x + 2) + c$.Step $4$: Compare with $e^x f(x) + c$ to find $f(x) = x^2 - 2x + 2$.Step $5$: Find the minimum value of $f(x) = x^2 - 2x + 2$.$f'(x) = 2x - 2$. Setting $f'(x) = 0$ gives $x = 1$.$f(1) = (1)^2 - 2(1) + 2 = 1 - 2 + 2 = 1$.

If $\int x^2 \cdot e^x dx = e^x f(x) + c$, then the minimum value of $f(x)$ is ...

Similar Questions

$\int \cos^{-1} x \, dx =$

$\int (\log x)^3 x^5 dx = $

Prove that $\int_{0}^{1} x e^{x} d x = 1$.

$\int \frac{x^2 \operatorname{Tan}^{-1} x}{(1+x^2)^2} dx =$

$\int x^3(\log x)^2 \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module