Chapter 11.2, Problem 29E

Solution

To find:the definite integral on the basis of given fact.

The definite integral of ${\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx}$ is $2$ .

Given data:

The area enclosed between the x-axis and one arch of the sine curve is $2$ .

The definite integral to compute is ${\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx}$.

Formula used:

The definite integral of function $f$ over $\left[a,b\right]$ denoted ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ is given by ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\left(x_i\right)}\Delta x$ provided the limit exists. If the limit exists, $f$ is integrable.

Calculation:

The definite integral is ${\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx}$.

Plot the graph of $y=\sin x$ over the interval $\left[0,\pi \right]$ .

  

From the graph, it can be observed that there is only one arch of $y=\sin x$ over the interval $\left[0,\pi \right]$ .

  ${\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx}=2$

Therefore, the definite integral of ${\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx}$ is $2$ .