Problem 26 - Calculus Early Transcendentals

Question

Evaluate the following integrals as they are written. $$\int_{0}^{\sqrt[3]{\pi / 2}} \int_{0}^{x} y \cos x^{3} d y d x$$

Step-by-Step Solution

Answer

Question: Evaluate the double integral given by the expression: $$\int_{0}^{\sqrt[3]{\pi / 2}} \int_{0}^{x} y \cos x^{3} d y d x$$ Answer: The result of the double integral is \(\frac{1}{6}\).

Step 1: Integrate with respect to y

To integrate the inner integral with respect to \(y\), observe that it is a simple product rule integration. $$\int_{0}^{x} y \cos x^{3} d y = \frac{1}{2} y^2\cos(x^3) \Big|_{0}^x$$ Now, after applying the limits we get: $$\frac{1}{2}x^2 \cos(x^{3})$$ For the outer integral, we will now integrate this expression with respect to \(x\).

Step 2: Integrate with respect to x

Now, we need to integrate the function \(\frac{1}{2}x^2 \cos(x^{3})\) with respect to \(x\) from \(0\) to \(\sqrt[3]{\pi / 2}\), but this function cannot be integrated directly using basic techniques. Therefore, we will use the substitution method. Let \(u = x^{3}\), so \(du = 3x^2 dx\). $$\int_{0}^{\sqrt[3]{\pi / 2}} \frac{1}{2} x^2 \cos(x^{3}) d x = \frac{1}{6} \int_{0}^{\pi / 2} \cos(u) du$$

Step 3: Evaluate the final integral

Now, we can easily integrate the last expression: $$\frac{1}{6}\int_{0}^{\pi / 2} \cos(u) du = \frac{1}{6}\Big[\sin(u) \Big|_{0}^{\pi/2}\Big] = \frac{1}{6}(\sin(\pi/2)-\sin(0))$$

Step 4: Simplify the expression

Now, let's simplify the expression: $$\frac{1}{6}(\sin(\pi/2)-\sin(0)) = \frac{1}{6}(1 - 0) = \frac{1}{6}$$ So, the result of the integral is \(\boxed{\frac{1}{6}}\).