SATHEE: Definite Integration Question 51

Definite Integration Question 51

Question 51

The value(s) of $\int_{0}^{1} \frac{x^{4} (1 - x)^{4}}{1 + x^{2}} d x$ is (are)

(2010) (a) $\frac{22}{7} - π$ (b) $\frac{2}{105}$ (c) 0 (d) $\frac{71}{15} - \frac{3 π}{2}$

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Solution:

Let $I = \int_{0}^{1} \frac{x^{4} (1 - x)^{4}}{1 + x^{2}} d x = \int_{0}^{1} \frac{(x^{4} - 1) (1 - x)^{4} + (1 - x)^{4}}{(1 + x^{2})} d x$

$\begin{aligned} = \int_{0}^{1} (x^{2} - 1) (1 - x)^{4} d x + \int_{0}^{1} \frac{{(1 + x^{2} - 2 x)}^{2}}{(1 + x^{2})} d x & = \int_{0}^{1} (x^{2} - 1) (1 - x)^{4} + (1 + x^{2}) - 4 x + \frac{4 x^{2}}{(1 + x^{2})} d x & = \int_{0}^{1} (x^{2} - 1) (1 - x)^{4} + (1 + x^{2}) - 4 x + 4 - \frac{4}{1 + x^{2}} d x & = \int_{0}^{1} x^{6} - 4 x^{5} + 5 x^{4} - 4 x^{2} + 4 - \frac{4}{1 + x^{2}} d x & = \frac{x^{7}}{7} - \frac{4 x^{6}}{6} + \frac{5 x^{5}}{5} - \frac{4 x^{3}}{3} + 4 x - 4 \tan^{- 1} x & = \frac{1}{7} - \frac{4}{6} + \frac{5}{5} - \frac{4}{3} + 4 - 4 \frac{π}{4} - 0 = \frac{22}{7} - π \end{aligned}$

Definite Integration Question 51

Question 51

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Definite Integration Question 51

Question 51

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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