The value of int_{-2012}^{2012} (sin(x^3) + x | vedclass

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

Question

(D) Let $I = \int_{-2012}^{2012} (\sin(x^3) + x^5 + 1) dx$. Using the property of linearity of integrals,we can write: $I = \int_{-2012}^{2012} \sin(x

Vedclass · Accepted Answer

$4024$

Vedclass · Answer

(D) Let $I = \int_{-2012}^{2012} (\sin(x^3) + x^5 + 1) dx$. Using the property of linearity of integrals,we can write: $I = \int_{-2012}^{2012} \sin(x^3) dx + \int_{-2012}^{2012} x^5 dx + \int_{-2012}^{2012} 1 dx$. Since $f(x) = \sin(x^3)$ and $g(x) = x^5$ are odd functions (i.e.,$f(-x) = -f(x)$),the integral of these functions over a symmetric interval $[-a, a]$ is $0$. Thus,$\int_{-2012}^{2012} \sin(x^3) dx = 0$ and $\int_{-2012}^{2012} x^5 dx = 0$. Therefore,$I = 0 + 0 + \int_{-2012}^{2012} 1 dx$. $I = [x]_{-2012}^{2012} = 2012 - (-2012) = 2012 + 2012 = 4024$.

The value of $\int_{-2012}^{2012} (\sin(x^3) + x^5 + 1) dx$ is

Similar Questions

Evaluate $\int_{0}^{\frac{\pi}{2}} \log \sin x \, dx$

$\int_{0}^{\frac{\pi}{2}} \frac{1-\cot x}{\operatorname{cosec} x+\cos x} d x=$

The value of $\int_0^{\pi / 2} \frac{(\cos x)^{\sin x}}{(\cos x)^{\sin x}+(\sin x)^{\cos x}} d x$ is

$\int_0^{\pi / 2} \frac{1}{1+\sqrt{\tan x}} d x=$

Let $f$ and $g$ be continuous functions on $[0, a]$ such that $f(x)=f(a-x)$ and $g(x)+g(a-x)=4$,then $\int_0^a f(x) g(x) d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module