left[ {sumlimits_{n = 1}^{10} {int_{ - 2n - 1 | vedclass

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

Question

(D) Let the given expression be $S = \sum\limits_{n = 1}^{10} \int_{ - 2n - 1}^{2n} \sin^{27}x \, dx + \sum\limits_{n = 1}^{10} \int_{2n}^{2n + 1} \si

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Let the given expression be $S = \sum\limits_{n = 1}^{10} \int_{ - 2n - 1}^{2n} \sin^{27}x \, dx + \sum\limits_{n = 1}^{10} \int_{2n}^{2n + 1} \sin^{27}x \, dx$. Consider the first summation: $I_1 = \sum\limits_{n = 1}^{10} \int_{ - 2n - 1}^{2n} \sin^{27}x \, dx$. Let $x = -t$,then $dx = -dt$. When $x = -2n-1$,$t = 2n+1$. When $x = 2n$,$t = -2n$. Thus,$\int_{ - 2n - 1}^{2n} \sin^{27}x \, dx = \int_{2n+1}^{-2n} \sin^{27}(-t) (-dt) = \int_{2n+1}^{-2n} -\sin^{27}t (-dt) = \int_{2n+1}^{-2n} \sin^{27}t \, dt = -\int_{-2n}^{2n+1} \sin^{27}t \, dt$. This approach is complex. Let's simplify the sum: $S = \sum\limits_{n = 1}^{10} \left( \int_{ - 2n - 1}^{2n} \sin^{27}x \, dx + \int_{2n}^{2n + 1} \sin^{27}x \, dx \right)$. By the property of definite integrals,$\int_{a}^{b} f(x) dx + \int_{b}^{c} f(x) dx = \int_{a}^{c} f(x) dx$. So,$S = \sum\limits_{n = 1}^{10} \int_{ - 2n - 1}^{2n + 1} \sin^{27}x \, dx$. Since $\sin^{27}x$ is an odd function,$\int_{-a}^{a} \sin^{27}x \, dx = 0$. Therefore,$\int_{ - (2n + 1)}^{(2n + 1)} \sin^{27}x \, dx = 0$ for each $n$. Thus,$S = \sum\limits_{n = 1}^{10} 0 = 0$.

$\left[ {\sum\limits_{n = 1}^{10} {\int_{ - 2n - 1}^{2n} {{{\sin }^{27}}x\,dx} } } \right] + \left[ {\sum\limits_{n = 1}^{10} {\int_{2n}^{2n + 1} {{{\sin }^{27}}x\,dx} } } \right]$ equals

Similar Questions

If ${I_n} = \int_0^{\pi /4} {{\tan ^n}\theta \,d\theta ,} $ then for any positive integer $n,$ the value of $n({I_{n - 1}} + {I_{n + 1}})$ is

$ \int_{0}^{\frac{\pi}{2}} \frac{\tan ^{7} x}{\cot ^{7} x+\tan ^{7} x} d x $ is equal to

If $M = \int_{0}^{\pi / 2} \frac{\cos x}{x+2} dx$ and $N = \int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{(x+1)^{2}} dx$, then the value of $M-N$ is

If $g(x)$ is the inverse of $f(x)$ and $f(x)$ has domain $x \in [1, 5]$,where $f(1) = 2$ and $f(5) = 10$,then the value of $\int_{1}^{5} f(x) dx + \int_{2}^{10} g(y) dy$ is equal to:

$\int_0^1 {\log \sin \left( {\frac{\pi }{2}x} \right)} \,dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module