The value of int_{0}^{1} frac{x^{4}(1-x)^{4}} | vedclass

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

Question

(A) Let $I = \int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} dx$. Expanding $(1-x)^{4} = 1 - 4x + 6x^{2} - 4x^{3} + x^{4}$. So,$I = \int_{0}^{1} \frac{x^{

Vedclass · Accepted Answer

$\frac{22}{7} - \pi$

Vedclass · Answer

(A) Let $I = \int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} dx$. Expanding $(1-x)^{4} = 1 - 4x + 6x^{2} - 4x^{3} + x^{4}$. So,$I = \int_{0}^{1} \frac{x^{4}(1 - 4x + 6x^{2} - 4x^{3} + x^{4})}{1+x^{2}} dx$. Performing polynomial division of $x^{4}(x^{4} - 4x^{3} + 6x^{2} - 4x + 1)$ by $(x^{2} + 1)$: $x^{8} - 4x^{7} + 6x^{6} - 4x^{5} + x^{4} = (x^{2} + 1)(x^{6} - 4x^{5} + 5x^{4} - 4x^{2} + 4) - 4$. Thus,$\frac{x^{4}(1-x)^{4}}{1+x^{2}} = x^{6} - 4x^{5} + 5x^{4} - 4x^{2} + 4 - \frac{4}{1+x^{2}}$. Integrating term by term from $0$ to $1$: $I = \int_{0}^{1} (x^{6} - 4x^{5} + 5x^{4} - 4x^{2} + 4) dx - 4 \int_{0}^{1} \frac{1}{1+x^{2}} dx$. $I = [\frac{x^{7}}{7} - \frac{4x^{6}}{6} + \frac{5x^{5}}{5} - \frac{4x^{3}}{3} + 4x]_{0}^{1} - 4[	an^{-1}(x)]_{0}^{1}$. $I = (\frac{1}{7} - \frac{2}{3} + 1 - \frac{4}{3} + 4) - 4(\frac{\pi}{4})$. $I = (\frac{3 - 14 + 21 - 28 + 84}{21}) - \pi = \frac{66}{21} - \pi = \frac{22}{7} - \pi$.

The value of $\int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} dx$ is equal to

Similar Questions

$\int_0^{\pi / 4} \tan ^2(x) \, dx =$

The value of the integral $\int_{1}^{5}[|x-3|+|1-x|] dx$ is equal to

The function $L(x) = \int_1^x \frac{dt}{t}$ satisfies the equation

Let $f$ be a polynomial function such that $f(x^{2}+1)=x^{4}+5x^{2}+2$ for all $x \in \mathbb{R}.$ Then $\int_{0}^{3} f(x) dx$ is equal to

Let $[t]$ denote the largest integer less than or equal to $t$. If $\int_0^3 \left( [x^2] + [\frac{x^2}{2}] \right) dx = a + b\sqrt{2} - \sqrt{3} - \sqrt{5} + c\sqrt{6} - \sqrt{7}$,where $a, b, c \in \mathbb{Z}$,then $a + b + c$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module