#### To determine

(a)

**To find:**

The interval in which C is increasing

#### Answer

4n-1≤|x|≤4n+1 for positive integers and |x|≤1

#### Explanation

**Concept:**

To find the interval in which a function is increasing, recall that the function is increasing if the first derivative C'x>0

**Given:**

Cx=∫0xcos12πt2dt

**Calculation:**

The given integral is

Cx=∫0xcos12πt2dt

Applying fundamental theorem of calculus

ddxCx=ddx∫0xcos12πt2dt

C'x =cos12πx2 The function C is increasing when C' is positive.

Recall that cosθ is positive in the interval -π2,π2

But cosθ is the periodic function with period 2π, therefore, cosθ is positive when

2nπ-π2≤θ≤2nπ+π2

Here the angle θ=12πx2, therefore C'x is positive 2nπ-π2≤12πx2≤2nπ+π2

Simplifying this 4n-1≤ x2≤4n+1 for every integer n

Taking square root

4n-1≤|x|≤4n+1

Therefore Cx is increasing in the interval 4n-1≤|x|≤4n+1 for every positive integer n and when |x|≤1

**Conclusion:**

The intervals in which C is increasing are 4n-1≤|x|≤4n+1 and |x|≤1

#### To determine

b)

**To find:**

The intervals in which C is concave upward of function

#### Answer

-2, 0, 2, 2, -6, -2, 6, 22…

In general they are of the form (-2(2n+1),-2n), (2(2n-1),2n) where n is any non-negative integer.

#### Explanation

**Concept:**

The function is concave up if C''x>0 in that particular interval.

**Given:**

C'x =cos12πt2 from part (a)

**Calculation:**

To find the interval of concave up, the second derivative is required because the function is concave up if C''x>0 in that particular interval.

C'x =cos12πt2, it was calculated in part a)

Differentiate again

C''x =-(πx)sin12πx2

This function sin12πx2 is negative in the intervals (2n-1)π≤ 12πx2≤2nπ

where n is positive integer (since 12πx2 is always non-negative)

Simplifying this

2(2n-1)≤ x2≤4n

Taking square root

2(2n-1)≤x≤2n Now sin (-x) = -sin (x) Similarly when x is negative sin12πx2 is positive when

-2(2n+1)≤x≤-2n where n is any non-negative integer

The function C''x contains the factor –x other than sin12πx2. So for x<0 is C''x is positive when sin12πx2 is positive. And for x>0C''x is positive when sin12πx2 is negative

Therefore, the function is concave up for x<0 when x is in (-2(2n+1),-2n) and the function is concave up for x>0 when x is in (2(2n-1),2n) where n is any non-negative integer.

Hence, the intervals of concave up for function Cx are

-2, 0, 2, 2, -6, -2, 6, 22....

**Conclusion:**

The intervals of concave up are -2, 0, 2, 2, -6, -2, 6, 22...

In general they are of the form (-2(2n+1),-2n), (2(2n-1),2n) where n is any non-negative integer.

#### To determine

c)

**To answer:**

∫0xcos12πt2dt=0.7 correct up to two decimal places by using the graph.

#### Answer

0.76 and 1.22

#### Explanation

**Given:**

∫0xcos12πt2dt=0.7

**Calculation:**

Now sketch the graph of ∫0xcos12πt2dt and the line y=0.7

From the graph, the solution of the function ∫0xcos12πt2dt=0.7 is the point where the line y=0.7 intersects the graph of ∫0xcos12πt2dt, and it is seen that the points are 0.76 and 1.22.

**Conclusion:**

The solution of the given expression is t=0.76 and t=1.22

#### To determine

d)

**To answer:**

To plot the graph of C and S to compare

#### Answer

#### Explanation

**1) Calculation:**

Plot the graphs for Cx=∫0xcos12πt2dt and Sx=∫0xsin12πt2dt

From the graph, we observe that both the curves converge to 12 as x→∞

**Conclusion:**

The graph of the functions are