#### To determine

**To find:**

The intervalwhere the function is increasing.

#### Answer

(-1, 1)

#### Explanation

**1) Concept:**

i) i) Fundamental Theorem of calculus. If f(x) is continuous on [a, b] and

gx=∫axf(t) dt

then g'x=f(x).

ii) Increasing test:

If g'x>0 on an interval, then g is increasing on that interval.

If g'x<0 on an interval, then g is decreasing on that interval.

**2) Given:**

fx=∫0x(1-t2)cos2t dt

**3) Calculation:**

Here,

fx=∫0x(1-t2)cos2t dt

Differentiating on both sides

f'x=ddx∫0x(1-t2)cos2t dt

By using the fundamental theorem

f'x=(1-x2)·cos2x

Therefore,

f'x=(1-x2)·cos2x

Now find x such that f'x=0 .

1-x2·cos2x=0

Simplifying,

1-x2=0

x2=1

x=-1, 1

These are the critical numbers of f, and they divide the domain into three intervals. To determine if f'x is negative or positive for each interval from the signs of the two factors of f'x namely, we shall use the following table.

Interval |
1-x2 |
cos2x |
f'x |
f |

x<-1 |
- |
+ |
- |
decreasing on (-∞,-1) |

-1<x<1 |
+ |
+ |
+ |
increasing on (-1,1) |

x>1 |
- |
+ |
- |
decreasing on (1,∞) |

The graph of f'(x) between interval (-1, 1) is as shown below,

Therefore, function f(x) is increasing on interval (-1, 1)

**Conclusion:**

Therefore, function is increasing in the interval (-1, 1)