Problem 79 - Calculus Early Transcendentals

Question

Find the area of the following regions. The region bounded by the graph of \(f(x)=x \sin x^{2}\) and the \(x\) -axis between \(x=0\) and \(x=\sqrt{\pi}\)

Step-by-Step Solution

Answer

Question: Find the area of the region bounded by the graph of the function \(f(x) = x \sin x^2\) and the x-axis between the values \(x=0\) and \(x=\sqrt{\pi}\). Answer: The area of the region is 1 square unit.

Step 1: Write down the integral expression for the area

To find the area of the region bounded by the function \(f(x)=x\sin x^2\) and the x-axis from \(x=0\) to \(x=\sqrt{\pi}\), we need to compute the definite integral: \(\int_{0}^{\sqrt{\pi}} x \sin x^2\, dx\)

Step 2: Use integration by substitution method

Integration by substitution (also known as the substitution rule or u-substitution) can be used to simplify the integration process. Let’s choose a substitution as: \(u = x^2\) Now, differentiate both sides with respect to x: \(\frac{du}{dx} = 2x\) Which can be rearranged to: \(dx = \frac{du}{2x}\) We also need to change the limits of integration according to the substitution: For \(x=0\), \(u=0^2=0\) For \(x=\sqrt{\pi}\), \(u=(\sqrt{\pi})^2=\pi\) The integral expression now becomes: \(\int_{0}^{\pi} x \sin u \cdot \frac{du}{2x}\) Note the \(x\)'s cancel out: \(\int_{0}^{\pi} \frac{\sin u}{2} du\)

Step 3: Integrate the new expression

Now, we can integrate our new expression: \(\int_{0}^{\pi} \frac{\sin u}{2} du = \left[-\frac{\cos u}{2}\right]_0^\pi\)

Step 4: Evaluate the integral

To find the area, we need to evaluate the integral at the limits: \(-\frac{\cos \pi}{2} - -\frac{\cos 0}{2} = \frac{1}{2} + \frac{1}{2} = 1\) Hence, the area of the region bounded by the graph of \(f(x)=x \sin x^2\) and the \(x\) -axis between \(x=0\) and \(x=\sqrt{\pi}\) is equal to 1 square unit.