The value of int_{-1}^{1} x^{2} e^{[x^{3}]} d | vedclass

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

Question

(C) Let $I = \int_{-1}^{1} x^{2} e^{[x^{3}]} dx$.We split the integral at $x = 0$:$I = \int_{-1}^{0} x^{2} e^{[x^{3}]} dx + \int_{0}^{1} x^{2} e^{[x^{

Vedclass · Accepted Answer

$\frac{e+1}{3e}$

Vedclass · Answer

(C) Let $I = \int_{-1}^{1} x^{2} e^{[x^{3}]} dx$.We split the integral at $x = 0$:$I = \int_{-1}^{0} x^{2} e^{[x^{3}]} dx + \int_{0}^{1} x^{2} e^{[x^{3}]} dx$.For $x \in [-1, 0)$,$x^{3} \in [-1, 0)$,so $[x^{3}] = -1$.For $x \in [0, 1)$,$x^{3} \in [0, 1)$,so $[x^{3}] = 0$.Substituting these values:$I = \int_{-1}^{0} x^{2} e^{-1} dx + \int_{0}^{1} x^{2} e^{0} dx$$I = \frac{1}{e} \int_{-1}^{0} x^{2} dx + \int_{0}^{1} x^{2} dx$$I = \frac{1}{e} \left[ \frac{x^{3}}{3} \right]_{-1}^{0} + \left[ \frac{x^{3}}{3} \right]_{0}^{1}$$I = \frac{1}{e} \left( 0 - \left( -\frac{1}{3} \right) \right) + \left( \frac{1}{3} - 0 \right)$$I = \frac{1}{3e} + \frac{1}{3} = \frac{1+e}{3e}$.

The value of $\int_{-1}^{1} x^{2} e^{[x^{3}]} dx$,where $[t]$ denotes the greatest integer $\leq t$,is

Similar Questions

$\int_2^4 \frac{\log x^2}{\log x^2+\log (36-12x+x^2)} dx$ is equal to

If $f(x) = \frac{e^x}{1+e^x}$, $l_1 = \int_{f(-a)}^{f(a)} x g(x(1-x)) dx$ and $l_2 = \int_{f(-a)}^{f(a)} g(x(1-x)) dx$, then the value of $\frac{l_2}{l_1}$ is

$\int_{0}^{a} (a-x)^{\frac{3}{2}} x^{2} dx =$

$\int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin x + \cos x} dx = $

For any integer $n,$ the integral $\int_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}(2n + 1)x\,dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module