Question

The accompanying figure shows four regions bounded by the graph of $y=x\sin x:R_1,R_2,R_3,$ and $R_4,$ whose areas are $1,\pi -1$ a $+1,$ and $2\pi -1,$ respectively. (We verify these results later in the text.) Use this information to evaluate the following integrals. $${\int }_0^{\pi }x\sin xdx$$

Short answer

** Answer: $2\pi -2$

Step-by-step solution

Identify intervals corresponding to regions

We will split the integral into 4 parts, one for each region. $[0,\frac{\pi }{2}]$ corresponds to regions $R_1$ and $R_2$ and $[\frac{\pi }{2},\pi ]$ corresponds to regions $R_3$ and $R_4$.

Use given areas to determine integrals

For the first half, we can write the integral as: $${\int }_0^{\frac{\pi }{2}}x\sin xdx=|-{\int }_0^{\frac{\pi }{2}}x\sin xdx|=Area(R_1)-Area(R_2)=1-(\pi -1)=2-\pi$$ For the second half, we have: $${\int }_{\frac{\pi }{2}}^{\pi }x\sin xdx=Area(R_3)+Area(R_4)=(\pi +1)+(2\pi -1)=3\pi$$

Combine the integrals

Finally, combine the results of Step 2 to find the integral over the given range $[0,\pi ]$: $${\int }_0^{\pi }x\sin xdx=({\int }_0^{\frac{\pi }{2}}x\sin xdx)+({\int }_{\frac{\pi }{2}}^{\pi }x\sin xdx)=(2-\pi )+(3\pi )=2\pi -2$$

Key concepts

Definite Integrals

A definite integral is a crucial concept in integral calculus. It helps us calculate the accumulation of quantities, like finding the area under a curve. Unlike indefinite integrals, which provide a general form of the antiderivative, definite integrals calculate a specific numerical value. This value represents the total change over an interval.Key facts about definite integrals include:

• The limits of integration determine the interval over which you are finding the accumulation.
• The notation ${\int }_a^{b}f(x)dx$ represents the definite integral of the function $f(x)$ from $a$ to $b$.
• They are used in various applications including physics, engineering, and economics.

In the exercise provided, definite integrals are used to calculate the areas of specific regions under the curve. By breaking down the problem into identifiable parts, definite integrals help with solving complex integrals by using known area values within given intervals.

Area Under a Curve

The concept of finding the area under a curve is closely tied to definite integrals. When you compute a definite integral, you are essentially calculating the area between the curve of the function and the x-axis, within the limits of integration.Here's how it works:

• For a curve defined by a function $y=f(x)$, the area from $x=a$ to $x=b$ can be found using ${\int }_a^{b}f(x)dx$.
• If the function is above the x-axis, the integral yields a positive area; if below, it gives a negative area.
• Complex curves can be split into simpler sections to determine the area of each individually, as shown in the solution steps of the problem.

In the given problem, areas of regions under the curve $y=x\sin x$ were predefined. These predefined areas helped directly compute the integrated value by identifying parts of interest under specific intervals of the function.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, play a major role in calculus problems. They describe oscillating systems and periodic phenomena-common in physics and engineering.Let's examine some special properties of trigonometric functions:

• The sine function, $\sin (x)$, oscillates between -1 and 1.
• These functions are periodic, repeating every $2\pi$ for sine and cosine.
• They are often involved in integrals to model real-world periodic behavior.

In this exercise, $y=x\sin x$ combines a linear function, $x$, with a trigonometric one, $\sin x$. This combination results in a wave-like curve whose areas under certain intervals are computed. Familiarity with the shape and properties of trigonometric functions helps understand how such curves behave across spaces and intervals.