Question

\( \int_{-1}^{4} \sqrt{\frac{4-x}{x+1}} \ dx = \)

• A. 0
• B. \( \frac{\pi}{2} \)
• C. \( \frac{3\pi}{2} \)
• D. \( \frac{5\pi}{2} \)

Hint:

Integrals of the form \( \int \sqrt{\frac{a-x}{x-b}} dx \) or \( \int \sqrt{\frac{a-x}{x+b}} dx \) often simplify with a trigonometric substitution. Here \( \int \sqrt{\frac{A-x}{x-B}} dx \), let \( x = B\cos^2\theta + A\sin^2\theta \). For \( \int \sqrt{\frac{a-x}{x+b}} \ dx \), complete the square in \( (a-x)(x+b) \) to get \( \sqrt{R^2 - (x-h)^2} \) form, then use \( x-h = R\sin\theta \). The integral \( \int \frac{P(x)}{\sqrt{ax^2+bx+c}} dx \) can often be split into \( K \int \frac{2ax+b}{\sqrt{ax^2+bx+c}} dx + L \int \frac{1}{\sqrt{ax^2+bx+c}} dx \). The form \( \int_{-a}^{a} \sqrt{a^2-x^2} dx \) is half the area of a circle, \( \frac{1}{2}\pi a^2 \). This integral doesn't immediately simplify to that.

Answer 1

The Correct Option is D

Let \( I = \int_{-1}^{4} \sqrt{\frac{4-x}{x+1}} \ dx \).
To rationalize, multiply numerator and denominator by \( \sqrt{4-x} \): \[ I = \int_{-1}^{4} \frac{4-x}{\sqrt{(x+1)(4-x)}} \ dx = \int_{-1}^{4} \frac{4-x}{\sqrt{4x-x^2+4-x}} \ dx = \int_{-1}^{4} \frac{4-x}{\sqrt{-x^2+3x+4}} \ dx \] Complete the square for the quadratic in the denominator: \( -x^2+3x+4 = -(x^2-3x-4) = -\left( \left(x-\frac{3}{2}\right)^2 - \frac{9}{4} - 4 \right) = -\left( \left(x-\frac{3}{2}\right)^2 - \frac{25}{4} \right) = \frac{25}{4} - \left(x-\frac{3}{2}\right)^2 \).
So, \( I = \int_{-1}^{4} \frac{4-x}{\sqrt{\frac{25}{4} - (x-\frac{3}{2})^2}} \ dx \).
Let \( x-\frac{3}{2} = \frac{5}{2}\sin\theta \).
Then \( dx = \frac{5}{2}\cos\theta d\theta \).
Also, \( 4-x = 4 - (\frac{3}{2}+\frac{5}{2}\sin\theta) = \frac{8-3-5\sin\theta}{2} = \frac{5-5\sin\theta}{2} = \frac{5}{2}(1-\sin\theta) \).
The denominator becomes \( \sqrt{\frac{25}{4} - \frac{25}{4}\sin^2\theta} = \sqrt{\frac{25}{4}\cos^2\theta} = \frac{5}{2}\cos\theta \).
Limits of integration: When \( x=-1 \): \( -1-\frac{3}{2} = \frac{5}{2}\sin\theta \implies -\frac{5}{2} = \frac{5}{2}\sin\theta \implies \sin\theta = -1 \implies \theta = -\pi/2 \).
When \( x=4 \): \( 4-\frac{3}{2} = \frac{5}{2}\sin\theta \implies \frac{5}{2} = \frac{5}{2}\sin\theta \implies \sin\theta = 1 \implies \theta = \pi/2 \).
\[ I = \int_{-\pi/2}^{\pi/2} \frac{\frac{5}{2}(1-\sin\theta)}{\frac{5}{2}\cos\theta} \cdot \frac{5}{2}\cos\theta d\theta \] \[ I = \int_{-\pi/2}^{\pi/2} \frac{5}{2}(1-\sin\theta) d\theta = \frac{5}{2} \int_{-\pi/2}^{\pi/2} (1-\sin\theta) d\theta \] \[ I = \frac{5}{2} [\theta + \cos\theta]_{-\pi/2}^{\pi/2} \] \[ I = \frac{5}{2} \left[ \left(\frac{\pi}{2} + \cos\frac{\pi}{2}\right) - \left(-\frac{\pi}{2} + \cos\left(-\frac{\pi}{2}\right)\right) \right] \] \[ I = \frac{5}{2} \left[ \left(\frac{\pi}{2} + 0\right) - \left(-\frac{\pi}{2} + 0\right) \right] = \frac{5}{2} \left[ \frac{\pi}{2} - (-\frac{\pi}{2}) \right] = \frac{5}{2} \left[ \frac{\pi}{2} + \frac{\pi}{2} \right] \] \[ I = \frac{5}{2} [\pi] = \frac{5\pi}{2} \] This matches option (4).