#### To determine

**To sketch and find:**

The area represented by g(x) and find the derivative by using Fundamental Theorem of Calculus.

#### Answer

By using Fundamental Theorem of Calculus part a) g'x=2+sinx

By using Fundamental Theorem of Calculus part b), g'x=2+sinx

#### Explanation

**1) Concept:**

The Fundamental Theorem of Calculus-Suppose f is continuous on [a, b] then

a) If gx=∫axftdt, then g'=fx

b) ∫abfxdx=Fb-F(a), F is an Antiderivative of f i.e F'=f

**2) Given:**

gx=∫0x(2+sint)dt

3) **Calculation:**

The area represented by gx=∫0x(2+sint)dt is the area between graph of fx=2+sinx and x axis from x = 0 to x = x

Now,

gx=∫0x(2+sint)dt

Take derivative of both sides

g'x=ddx∫0x(2+sint)dt

Use Fundamental Theorem of Calculus part a) with ft=2+sint to get

g'(x)=2+sinx

Therefore, g'(x)=2+sinx

Now by using part b) of Fundamental Theorem of Calculus,

gx=∫0x(2+sint)dt

Here ft=2+sint

The anti derivative of f is Gt=2t-cost

Use Fundamental Theorem of Calculus part b) to get

gx=∫0x(2+sint)dt=Gx-G(0)

Use G(t)=2t-cost so G(x)=2x-cosx and G(0)=2*0-cos(0)

So gx=2x-cosx-(-cos0)

Simplifying

gx=2x-cosx+1

Taking derivative

g'(x)=2(1)-(-sinx)-0

Simplifying

g'(x)=2+sinx

Therefore, g'(x)=2+sinx

**Conclusion:**

Therefore, by using Fundamental Theorem of Calculus part a) g'x=2+sinx

and by using Fundamental Theorem of Calculus part b), g'x=2+sinx