int_{0}^{2} x e^{x} dx = | vedclass

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

Question

(A) To evaluate the integral $\int_{0}^{2} x e^{x} dx$,we use the method of integration by parts: $\int u dv = uv - \int v du$.Let $u = x$ and $dv = e

Vedclass · Accepted Answer

$e^{2} + 1$

Vedclass · Answer

(A) To evaluate the integral $\int_{0}^{2} x e^{x} dx$,we use the method of integration by parts: $\int u dv = uv - \int v du$.Let $u = x$ and $dv = e^{x} dx$. Then $du = dx$ and $v = e^{x}$.Applying the formula: $\int x e^{x} dx = x e^{x} - \int e^{x} dx = x e^{x} - e^{x}$.Now,applying the limits from $0$ to $2$:$\left[ x e^{x} - e^{x} \right]_{0}^{2} = (2 e^{2} - e^{2}) - (0 \cdot e^{0} - e^{0})$.$= (e^{2}) - (0 - 1) = e^{2} + 1$.

$\int_{0}^{2} x e^{x} dx =$

Similar Questions

The value of the definite integral $\int\limits_{\infty}^{0} \frac{z e^{-z}}{\sqrt{1-e^{-2z}}} \, dz$ is:

The value of $\int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} dx$ is equal to

The function $L(x) = \int_1^x \frac{dt}{t}$ satisfies the equation

The value of $I = \int_0^{\pi /2} \frac{(\sin x + \cos x)^2}{\sqrt{1 + \sin 2x}} dx$ is

$\int_{-1}^1 \frac{\cosh x}{1+e^{2 x}} d x$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module