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Chapter 6: Problem 36

Evaluate the integral using (a) the given integration limits and (b) the limits obtained by trigonometric substitution. $$ \int_{0}^{3 / 5} \sqrt{9-25 x^{2}} d x $$

Short Answer

Expert verified
The solution to the integral \(\int_{0}^{3 / 5} \sqrt{9-25 x^{2}} d x\) with the given limits is 3

Step by step solution

01

Initial Integral Evaluation

Begin the problem using the given integration limits from 0 to 3/5. The integral to be evaluated is \[ \int_{0}^{3 / 5} \sqrt{9-25 x^{2}} d x \]
02

Trigonometric Substitution

The expression inside the square root, \(9 - 25x^2\), resembles difference of squares. Take \(x = \frac{3}{5} sin(t)\) as a substitution. This automatically changes the limits of integration due to \(t = arcsin(\frac{x}{3/5})\). For \(x = 0, t = 0\) and for \(x = 3/5, t = 蟺/2\). Hence, the integral becomes \[\int_{0}^{\pi/2} \sqrt{9 - 25(\frac{3}{5} sin(t))^2} * \frac{3}{5} cos(t) dt = \int_{0}^{\pi/2} 3 cos(t) dt\]
03

Evaluate the New Integral

Now the integral can be evaluated easily as the antiderivative of cosine is sine. Therefore we get \[3[\sin(t)]_{0}^{\pi/2} = 3(1 - 0) = 3\]

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