int_0^2 x^2(2-x)^5 d x= | vedclass

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

Question

(C) To evaluate the integral $I = \int_0^2 x^2(2-x)^5 dx$,we use the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$. Substituting $x = 2-t$,we get $

Vedclass · Accepted Answer

$\frac{32}{21}$

Vedclass · Answer

(C) To evaluate the integral $I = \int_0^2 x^2(2-x)^5 dx$,we use the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$. Substituting $x = 2-t$,we get $dx = -dt$. When $x=0, t=2$ and when $x=2, t=0$. $I = \int_2^0 (2-t)^2(t)^5 (-dt) = \int_0^2 (4 - 4t + t^2)t^5 dt$. $I = \int_0^2 (4t^5 - 4t^6 + t^7) dt$. Integrating term by term: $I = [\frac{4t^6}{6} - \frac{4t^7}{7} + \frac{t^8}{8}]_0^2$. $I = [\frac{2(2^6)}{3} - \frac{4(2^7)}{7} + \frac{2^8}{8}] = [\frac{128}{3} - \frac{512}{7} + 32]$. Finding a common denominator of $21$: $I = \frac{896 - 1536 + 672}{21} = \frac{32}{21}$. Thus,the correct option is $C$.

$\int_0^2 x^2(2-x)^5 d x=$

Similar Questions

If $f(x)$ is an odd function of $x,$ then $\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {f(\cos x)\,dx} $ is equal to

$\int_{1}^{6\pi}([\sec^{-1}x]+[\cot^{-1}x])dx$ is equal to (where $[.]$ denotes the greatest integer function).

Given $\int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sin x + \cos x} = \ln 2$,then the value of the definite integral $\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1 + \sin x + \cos x} dx$ is equal to

$\int_{- 1 / 2}^{1 / 2} \{ [x] + \log (\frac{1 + x}{1 - x}) \} dx =$

$\int_{ - \pi }^{\pi } {\frac{{2x(1 + \sin x)}}{{1 + {{\cos }^2}x}}dx} $ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module