I = intlimits_0^1 sqrt[3]{2x^3 - 3x^2 - x + 1 | vedclass

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

Question

(B) Let $I = \int\limits_0^1 f(x) \, dx$,where $f(x) = \sqrt[3]{2x^3 - 3x^2 - x + 1}$.Using the property $\int\limits_a^b f(x) \, dx = \int\limits_a^b

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Let $I = \int\limits_0^1 f(x) \, dx$,where $f(x) = \sqrt[3]{2x^3 - 3x^2 - x + 1}$.Using the property $\int\limits_a^b f(x) \, dx = \int\limits_a^b f(a+b-x) \, dx$,we have $I = \int\limits_0^1 f(1-x) \, dx$.Calculate $f(1-x) = \sqrt[3]{2(1-x)^3 - 3(1-x)^2 - (1-x) + 1}$.$f(1-x) = \sqrt[3]{2(1 - 3x + 3x^2 - x^3) - 3(1 - 2x + x^2) - 1 + x + 1}$.$f(1-x) = \sqrt[3]{2 - 6x + 6x^2 - 2x^3 - 3 + 6x - 3x^2 - 1 + x + 1}$.$f(1-x) = \sqrt[3]{-2x^3 + 3x^2 + x - 1} = -\sqrt[3]{2x^3 - 3x^2 - x + 1} = -f(x)$.Therefore,$I = \int\limits_0^1 -f(x) \, dx = -I$.$2I = 0 \implies I = 0$.

$I = \int\limits_0^1 \sqrt[3]{2x^3 - 3x^2 - x + 1} \, dx$ is equal to

Similar Questions

Let $f:R \to R$ and $g:R \to R$ be continuous functions,then the value of the integral $\int_{-\pi/2}^{\pi/2} [f(x) + f(-x)][g(x) - g(-x)] \, dx$ is:

Let $A = \int\limits_0^1 \frac{e^t}{1 + t} \, dt$. Then $\int\limits_{a - 1}^a \frac{e^{-t}}{t - a - 1} \, dt$ has the value:

$\int_{ - 3}^3 {\frac{{{x^2}\sin 2x}}{{{x^2} + 1}}\,dx = } $

The value of $\int_{0}^{\frac{\pi}{2}} \ln \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ is

$\int_0^{\frac{\pi}{2}} \frac{x \tan x \sec^2 x}{\tan^4 x + 1} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module