_{-1}^{4} {4-x/x+1} dx = [AP EAPCET 22 May 2025 S2]

Correct Answer is : 5π25π/225π

Correct Answer is : 5π25π/225π

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Definite Integration

Section: Mathematics

Question : 11 of 54

Marks: +1, -0

\int\limits_{-1}^{4} \sqrt{\frac{4-x}{x+1}} dx =

Solution:

\; \int\limits_{-1}^{4} \sqrt{\;\frac{4-x}{x+1}} dx

\; \; \text{ Let } \; x = \frac{5}{2} \sin \;\theta + \frac{3}{2}

\; \Rightarrow dx = \frac{5}{2} \cos \;\theta d \theta

\; \Rightarrow 4 - x = 4 - \left( \frac{5}{2} \sin \;\theta + \frac{3}{2} \right)

\; = \frac{5}{2} (1 - \sin \;\theta)

\; \Rightarrow x + 1 = \frac{5}{2} \sin \;\theta + \frac{3}{2} + 1 = \frac{5}{2} (1 + \sin \;\theta)

\; \Rightarrow \; \; \frac{4-x}{x+1} = \frac{1 - \sin \;\theta}{1 + \sin \;\theta}

\; \Rightarrow \; \; \sqrt{\;\frac{4-x}{x+1}} = \sqrt{\;\frac{1 - \sin \;\theta}{1 + \sin \;\theta}} = \frac{1 - \sin \;\theta}{\cos \theta}

\; \int \sqrt{\;\frac{4-x}{x+1}} dx = \int \frac{1 - \sin \;\theta}{\cos \theta} \;\frac{5}{2} \cos \theta d \theta

\; = \frac{5}{2} \int (1 - \sin \;\theta) d \theta

\; = \frac{5}{2} (\theta + \cos \theta) + c

\; x = \frac{5}{2} \sin \;\theta + \frac{3}{2} \Rightarrow \sin \;\theta = \frac{2x - 3}{5}

\; \Rightarrow \cos \theta = \sqrt{1 - \sin^2 \theta} = \frac{2}{5} \sqrt{4 + 3x - x^2}

\; \int\limits_{-1}^{4} \sqrt{\;\frac{4-x}{x+1}} dx = \left[ \frac{5}{2} \sin^{-1} \left( \frac{2x - 3}{5} \right) + \sqrt{4 + 3x - x^2} \right]_{-1}^{4}

\; \; = \frac{5}{2} \cdot \frac{\pi}{2} - \frac{5}{2} \left( \frac{-\pi}{2} \right)

\; \; = \frac{5\pi}{4} + \frac{5\pi}{4} = \frac{5\pi}{2}

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