Problem 14 - College Algebra and Calculus: An Applied Approach

Question

Evaluate the double integral. $$ \int_{0}^{1} \int_{0}^{x} \sqrt{1-x^{2}} d y d x $$

Step-by-Step Solution

Answer

The value of the double integral is -1.

Step 1: Understand the Iterated Integral

An iterated integral like \(\int_{0}^{1} \int_{0}^{x} \sqrt{1-x^{2}} dy dx\) involves performing multiple integrations, one after the other. The inner integral is done first, with respect to \(y\), then the result is integrated with respect to \(x\). The limits of the integral for \(y\) are from 0 to \(x\), and for \(x\) are from 0 to 1.

Step 2: Perform the Inner Integral

The inner integral is \(\int_{0}^{x} \sqrt{1-x^{2}} dy\). The limits of integration are \(y = 0\) to \(y = x\). The function to integrate does not contain \(y\), so the integral simplifies to \([y\sqrt{1-x^{2}}]_{0}^{x}\), equal to \(x\sqrt{1-x^{2}} - 0\), which further simplifies to \(x\sqrt{1-x^{2}}\).

Step 3: Perform the Outer Integral

We now need to perform the outer integral, which is \(\int_{0}^{1} x\sqrt{1-x^{2}} dx\). Here, the suggestion is to use the substitution method. Let \(u = x^{2}\) (thus \(du = 2xdx\)), then the integral becomes \(-\frac{1}{2} \int_{0}^{1} \sqrt{1-u} du\), where \(\sqrt{1-u}\) is the integral of \(u^{-1/2}\). The antiderivative of \(u^{-1/2}\) is \(2u^{1/2}\), so the evaluation of the definite integral becomes \(-\frac{1}{2} [2u^{1/2}]_{0}^{1}\), which simplifies to \(-1 + 0\).