If int e^{x^2} cdot x^3 , dx = e^{x^2} f(x) + | vedclass

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

Question

(D) Let $I = \int e^{x^2} \cdot x^3 \, dx$. Substitute $t = x^2$,then $dt = 2x \, dx$,which implies $x \, dx = \frac{1}{2} dt$. Substituting these int

Vedclass · Accepted Answer

$\frac{x^2-1}{2}$

Vedclass · Answer

(D) Let $I = \int e^{x^2} \cdot x^3 \, dx$. Substitute $t = x^2$,then $dt = 2x \, dx$,which implies $x \, dx = \frac{1}{2} dt$. Substituting these into the integral,we get: $I = \frac{1}{2} \int e^t \cdot t \, dt$. Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = t$ and $dv = e^t \, dt$: $I = \frac{1}{2} (t e^t - \int e^t \, dt) = \frac{1}{2} (t e^t - e^t) + c$. $I = \frac{1}{2} e^t (t - 1) + c$. Substituting $t = x^2$ back,we get: $I = \frac{1}{2} e^{x^2} (x^2 - 1) + c$. Comparing this with the given form $e^{x^2} f(x) + c$,we find $f(x) = \frac{x^2 - 1}{2}$. Since $f(1) = \frac{1^2 - 1}{2} = 0$,the condition is satisfied.

If $\int e^{x^2} \cdot x^3 \, dx = e^{x^2} f(x) + c$ and $f(1) = 0$ (where $c$ is a constant of integration),then the value of $f(x)$ is

Similar Questions

If $f(x) = \frac{\sin^{-1} x}{\sqrt{1-x^2}}$ and $g(x) = e^{\sin^{-1} x}$, then the value of $\int f(x)g(x) dx = \dots$

Evaluate the integral: $\int \sin(\log x) dx$

$\int \sin(\log x) \, dx = $

$\int \frac{x^3}{\sqrt{1+x^2}} dx$ is equal to

Integrate the function: $\tan ^{-1} \sqrt{\frac{1-x}{1+x}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module