The value(s) of int_0^1 frac{x^4(1-x)^4}{1+x^ | vedclass

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

Question

(A) We need to evaluate the integral $I = \int_0^1 \frac{x^4(1-x)^4}{1+x^2} d x$.First,expand the numerator: $x^4(1-x)^4 = x^4(1-4x+6x^2-4x^3+x^4) = x

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) We need to evaluate the integral $I = \int_0^1 \frac{x^4(1-x)^4}{1+x^2} d x$.First,expand the numerator: $x^4(1-x)^4 = x^4(1-4x+6x^2-4x^3+x^4) = x^8-4x^7+6x^6-4x^5+x^4$.Now,perform polynomial division of $x^8-4x^7+6x^6-4x^5+x^4$ 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:$I = \int_0^1 (x^6-4x^5+5x^4-4x^2+4) dx - \int_0^1 \frac{4}{1+x^2} dx$.$I = \left[ \frac{x^7}{7} - \frac{4x^6}{6} + \frac{5x^5}{5} - \frac{4x^3}{3} + 4x ight]_0^1 - 4[	an^{-1}(x)]_0^1$.$I = \left( \frac{1}{7} - \frac{2}{3} + 1 - \frac{4}{3} + 4 ight) - 4(\frac{\pi}{4})$.$I = \left( \frac{1}{7} - 2 + 5 ight) - \pi = \frac{1}{7} + 3 - \pi = \frac{22}{7} - \pi$.

The value$(s)$ of $\int_0^1 \frac{x^4(1-x)^4}{1+x^2} d x$ is (are)

Similar Questions

The approximate value of $\int_{1}^{9} x^2 dx$ by using the trapezoidal rule with $4$ equal intervals is:

If $\int_{1}^{k}(3x^{2}+2x+1)dx=11$,then $k=$

$\int_0^1(\sqrt{10})^{2x} dx=$

$\int_{0}^{\pi/4} \sqrt{1+\sin 2x} dx = \rule{1cm}{0.15mm}$

Evaluate the definite integral: $\int_{0}^{a} \sqrt{\frac{x}{a-x}} \, dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module