If int f'(x) cdot e^{x^2} dx = (x - 1) cdot e | vedclass

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

Question

(D) Given $\int f'(x) e^{x^2} dx = (x - 1) e^{x^2} + k$.Differentiating both sides with respect to $x$:$f'(x) e^{x^2} = \frac{d}{dx} [(x - 1) e^{x^2}]

Vedclass · Accepted Answer

$\frac{2x^3}{3} - x^2 + x + c$, where $c$ is the constant of integration.

Vedclass · Answer

(D) Given $\int f'(x) e^{x^2} dx = (x - 1) e^{x^2} + k$.Differentiating both sides with respect to $x$:$f'(x) e^{x^2} = \frac{d}{dx} [(x - 1) e^{x^2}]$$f'(x) e^{x^2} = (1) e^{x^2} + (x - 1) e^{x^2} (2x)$$f'(x) e^{x^2} = e^{x^2} [1 + 2x^2 - 2x]$$f'(x) = 2x^2 - 2x + 1$Integrating $f'(x)$ with respect to $x$:$f(x) = \int (2x^2 - 2x + 1) dx$$f(x) = \frac{2x^3}{3} - x^2 + x + c$Thus, option $D$ is correct.

If $\int f'(x) \cdot e^{x^2} dx = (x - 1) \cdot e^{x^2} + k$, where $k$ is the constant of integration, then $f(x) = \dots$

Similar Questions

If $\int \frac{x^4+1}{x^2+1} dx = Ax^3 + Bx^2 + Cx + D \tan^{-1} x + E$,then $A+B+C+D=$

$\int \frac{d x}{\cos (x+4) \cos (x+2)}$ is equal to

$\int \frac{x^8-9 x^2+18}{x^4-3 x^2+3} d x=$

Find the following integral: $\int(1-x) \sqrt{x} \, dx$

$\int \tan^2 x \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module