# Use F(x) = \int_x^9 (\frac 2{1 + t^2} - 10)dt to answer the questions below: a. Evaluate the...

## Question:

Use {eq}F(x) = \int_x^9 (\frac 2{1 + t^2} - 10)dt {/eq} to answer the questions below:

a. Evaluate the integral to find a formula for F(x).

b. Find F'(x) by computing your formula from part a.

c. Find F'(x) using the Second Fundamental Theorem of Calculus.

## !!!Fundamental Theorem of Calculus:

The differentiation of any integration of any function which is an actual function. For example, take the derivative of integration of function:

{eq}\displaystyle \frac{d}{dy} \int_b^y g(t) dt = g(x). {/eq}

Commonly used applications for the integration:

1. The sum rule: {eq}\displaystyle \int f\left(u\right)\pm g\left(u\right)du=\int f\left(u\right)du\pm \int g\left(u\right)du. {/eq}

2. Eject the constant out: {eq}\displaystyle \int a\cdot f\left(u\right)du=a\cdot \int f\left(u\right)du. {/eq}

3. Common integration: {eq}\displaystyle \int \frac{1}{1+t^2}dt=\arctan \left(t\right). {/eq}

4. Integration of a constant: {eq}\displaystyle \int a du=au. {/eq}

5. The sum/difference rule: {eq}\displaystyle \left(f\pm g\right)'=f'\pm g'. {/eq}

6. Derivative of a constant: {eq}\displaystyle \frac{d}{dx}\left(a\right)=0. {/eq}

7. Common derivative: {eq}\displaystyle \frac{d}{dx}\left(\arctan \left(x\right)\right)=\frac{1}{x^2+1}. {/eq}

8. Common derivative: {eq}\displaystyle \frac{d}{dx}\left(x\right)=1. {/eq}

## Answer and Explanation:

Here, we have to use the second fundamental theorem of calculus for $$\begin{align*} \displaystyle F'(x) &= \frac{d}{dx} \int_x^9 (\frac 2{1 + t^2} - 10)dt\\ \displaystyle &= \frac{d}{dx} I. \end{align*} $$

Now, we are going to find I.

{eq}\displaystyle I = \int_x^9 (\frac 2{1 + t^2} - 10)dt {/eq}

Apply the sum rule.

{eq}\displaystyle = \int _x^9\frac{2}{1+t^2}dt-\int _x^910dt {/eq}

Eject the constant out.

{eq}\displaystyle = 2 \cdot \int _x^9\frac{1}{1+t^2}dt-\int _x^910dt {/eq}

Apply the common integration and integration of a constant.

{eq}\displaystyle = 2\left[\arctan \left(t\right)\right]^9_x-\left[10t\right]^9_x {/eq}

Compute the boundaries.

{eq}\begin{align*} \displaystyle &= 2\left(\arctan \left(9\right)-\arctan \left(x\right)\right)-\left(90-10x\right)\\ \displaystyle &= 2\arctan \left(9\right)-2\arctan \left(x\right)-90+10x. \end{align*} {/eq}

Put in the value of I.

$$\displaystyle = \frac{d}{dx} 2\arctan \left(9\right)-2\arctan \left(x\right)-90+10x $$

Apply the sum/difference rule.

$$\displaystyle = \frac{d}{dx}\left(2\arctan \left(9\right)\right)-\frac{d}{dx}\left(2\arctan \left(x\right)\right)-\frac{d}{dx}\left(90\right)+\frac{d}{dx}\left(10x\right) $$

Eject the constant out.

$$\displaystyle = \frac{d}{dx}\left(2\arctan \left(9\right)\right)-2\frac{d}{dx}\left(\arctan \left(x\right)\right)-\frac{d}{dx}\left(90\right)+10\frac{d}{dx}\left(x\right) $$

Apply the common derivative.

$$\displaystyle = 0-\frac{2}{x^2+1}-0+10 $$

Simplify:

$$\displaystyle = -\frac{2}{x^2+1}+10. $$