Question

Compute the definite integrals. Use a graphing utility to confirm your answers. $$ \int_{0}^{\pi / 2} x^{2} \sin x d x \text { (Express the answer in exact form.) } $$

Short answer

The integral evaluates to \(\pi\).

Step-by-step solution

Choose Integration Technique

For the integral \( \int_{0}^{\pi / 2} x^{2} \sin x \, dx \), notice that it involves a polynomial \( x^2 \) and a trigonometric function \( \sin x \). Integration by parts is a suitable method for solving this integral.

Apply Integration by Parts Formula

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Choose \( u = x^2 \) and \( dv = \sin x \, dx \). Then, compute \( du = 2x \, dx \) and \( v = -\cos x \) by integrating \( dv \).

Substitute into Integration by Parts Formula

Substitute the chosen \( u \), \( du \), \( v \), and \( dv \) into the formula: \( \int x^2 \sin x \, dx = -x^2 \cos x \bigg|_0^{\pi/2} + \int 2x \cos x \, dx \).

Apply Integration by Parts Again

For the integral \( \int 2x \cos x \, dx \), apply integration by parts again. Choose \( u = 2x \) and \( dv = \cos x \, dx \). Then, compute \( du = 2 \, dx \) and \( v = \sin x \). Substitute into integration by parts to find \( \int 2x \cos x \, dx = 2x \sin x \bigg|_0^{\pi/2} - 2 \int \sin x \, dx \).

Integrate and Simplify

Compute the remaining integral \( \int \sin x \, dx = -\cos x \). Substitute back to get \( -x^2 \cos x + (2x \sin x + 2 \cos x) \bigg|_0^{\pi/2} \). Evaluate this expression at the limits from 0 to \( \pi/2 \).

Evaluate the Expression at Limits

Evaluate this expression: At \( x = \frac{\pi}{2} \), you get the terms \( -\left(\frac{\pi}{2}\right)^2 \cdot 0 + (2 \cdot \frac{\pi}{2} \cdot 1 + 2 \cdot 0) = \pi \). At \( x = 0 \), you get 0 due to the multiplication by zero in each term. Hence, the solution is \( \pi \).

Confirm with Graphing Utility

Use a graphing utility to confirm the answer by plotting the integral of \( x^2 \sin x \) from 0 to \( \pi/2 \) and ensuring the area under the curve also computes to \( \pi \).

Key concepts

Definite Integrals

A definite integral allows us to find the area under the curve of a function over a specified interval. In this exercise, we are looking at the interval from 0 to \( \frac{\pi}{2} \). The result of a definite integral is a single number, which represents this area.

• The integral's limits, given here as 0 and \( \frac{\pi}{2} \), define the boundaries of the area to calculate.
• When evaluating a definite integral, apply the fundamental theorem of calculus. This involves finding the antiderivative and evaluating it at the upper and lower limits of integration.
• In this particular case, the integral of \( x^2 \sin x \) helps us solve the problem by breaking it down using specific integration techniques such as integration by parts.

After solving through analytic methods, it's beneficial to use tools like graphing utilities to confirm the result visually by the shape and area under the curve within these boundaries.

Trigonometric Functions

Trigonometric functions are key players in calculus and integration because of their periodic nature and unique properties. In our exercise, the function \( \sin x \) is crucial.

• \( \sin x \) is a basic trigonometric function that varies smoothly and predictably, with values oscillating between -1 and 1.
• When integrating functions involving \( \sin x \), you often use specific integral techniques to handle the product of trigonometric and other algebraic functions, such as polynomials.
• The integral of \( \sin x \) is \(-\cos x\), an important antiderivative commonly encountered during integration by parts.

The combination with polynomials (like \( x^2 \)) in integrals can look complex initially, but utilizing methods like integration by parts makes it manageable and offers insight into how different types of functions interact when integrated.

Polynomial Functions

Polynomial functions, like \( x^2 \) in this exercise, are foundational to calculus and integral calculus. Understanding them makes handling integrals easier.

• Polynomials are expressions consisting of variables raised to whole number powers, with coefficients multiplying these terms.
• The polynomial \( x^2 \) is straightforward to integrate by itself, resulting in \( \frac{1}{3}x^3 \) when using the basic integration rule for power functions.
• When paired with trigonometric functions, as we see here, it becomes necessary to use methods such as integration by parts to break down and integrate the combined function effectively.

Solving integrals involving polynomials like \( x^2 \) often requires strategic approaches, but mastering these techniques aids in solving more advanced calculus problems with ease.