Q. 51 - Calculus

Question

Use the Fundamental Theorem of Calculus to find the exact

values of each of the definite integrals in Exercises 19–64. Use

a graph to check your answer. (Hint: The integrands that involve

absolute values will have to be considered piecewise.)

${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sin x(1+\cos x)dx$

Step-by-Step Solution

Answer

${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sin x(1+\cos x)dx=\frac{3}{2}$.

Step 1. Given information.

A definite integral is given as ${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sin x(1+\cos x)dx$.

Step 2. Using the Fundamental theorem of Calculus.

We get

$$\begin{gathered} {\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sin x(1+\cos x)dx \\ ={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sin xdx+{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sin x\cos xdx \\ ={[-\cos x]}_0^{\frac{\mathrm{\pi }}{2}}+\frac{1}{2}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}2\sin x\cos xdx \\ =-\cos \frac{\mathrm{\pi }}{2}+\cos 0+\frac{1}{2}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sin \left(2x\right)dx \\ =-0+1+\frac{1}{2}{[-\frac{1}{2}\cos (2x\left)\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ =1-\frac{1}{4}[\mathrm{cos\pi }-\cos 0] \\ =1-\frac{1}{4}(-1-1) \\ =1+\frac{1}{2} \\ =\frac{3}{2} \end{gathered}$$

So the exact value of the given definite integral is $\frac{3}{2}$.

Step 3. The graph to verify the answer is

The solution is area under graph which is 

$$\begin{gathered} a=1.5 \\ =\frac{3}{2} \end{gathered}$$