Chapter 5.4, Problem 6E

Sketch the area represented by g(x). Then find g′(x) in two ways: (a) by using Part 1 of the Fundamental Theorem and (b) by evaluating the integral using Part 2 and then differentiating.

$g(x)={\displaystyle {\int }_0^x(2+\sin t)}dt$

To determine

To calculate: The $g^{\prime }\left(x\right)$, using part 1 of fundamental theorem of calculus and integral evaluation.

Answer to Problem 6E

The $g^{\prime }\left(x\right)$ value is $\underset{\_}{2+\sin x}$ by using part 1 of fundamental theorem of calculus method.

Explanation of Solution

Given information:

$g\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}\left(2+\sin t\right)dt}$

Calculation:

Show the integral function as shown below:

$g\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}\left(2+\sin t\right)dt}$ (1)

Here, $g\left(x\right)$ is function of x andt is a derivative.

Draw the area represented $g\left(x\right)$ as below;

Show the part 1 of fundamental theorem of calculus as shown below:

$g\left(x\right)={\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)dt}$ (2)

$g^{\prime }\left(x\right)=\frac{d}{dx}{\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)dt}$ (3)

Comparing Equation (1), (2) and (3).

$\begin{array}{c}g\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}\left(2+\sin t\right)dt}\\ g^{\prime }\left(x\right)=f\left(x\right)\\ =2+\sin x.\end{array}$

Therefore, get the value of $g^{\prime }\left(x\right)$ as $\underset{\_}{2+\sin x}$ by using the part 1 of fundamental theorem of calculus.

To determine

The value for $g^{\prime }\left(x\right)$ by using the integral evaluation method.

Answer to Problem 6E

The $g^{\prime }\left(x\right)$ value is $\underset{\_}{2+\sin x}$ by using the integral evaluation method.

Explanation of Solution

Given information:

$g\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}\left(2+\sin t\right)dt}$

Calculation:

Apply the integral evaluation method:

$\begin{array}{c}g\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}\left(2+\sin t\right)dt}\\ ={\left(2t-\cos t\right)}_0^x\\ =\left[\left(2x-\cos x\right)-\left(0-\cos 0\right)\right]\\ =2x-\cos x+1.\end{array}$

Differentiate with respect to x.

$\begin{array}{c}g\left(x\right)=2x-\cos x+1\\ g^{\prime }\left(x\right)=2\left(1\right)-\left(-\sin x\right)\\ =2+\sin x.\end{array}$

Therefore, get the value of $g^{\prime }\left(x\right)$ as $\underset{\_}{2+\sin x}$ using the integral method.