#### To determine

The given statement is true or false.

#### Answer

False

#### Explanation

Recall theFundamentaltheorem of calculus part 1:

If f is continuous on a,b, then the function g defined by gx= ∫axftdt a ≤x ≤b is continuous on a,b and differentiable on (a, b), and g'x=f(x)

Let fx=3x+2 for 0≤x≤1

Here the upper limit oftheintegration in variable x is not a constant number.

If both limits on integration are constant numbers, then the integral is constant. Since derivative of constant is zero, answer must be zero.

Consider the example,

∫013x+2dx

fx=3x+2 is continuous on [0, 1] since it is polynomial.

∫013x+2dx=3x22+2x01

By using the fundamental theorem of calculus,

3x22+2x01=12·32+2-0

By simplifying,

∫013x+2dx=72

ddx∫013x+2dx=ddx72=0

Here,

ddx∫abfxdx=ddx∫013x+2dx=0≠fx=3x+2

So, the given statement is false.

**Conclusion:**

The given statement is false.