#### To determine

(a)

**To draw:**

Graph of integral Si

#### Answer

#### Explanation

**Concept:**

Use Computer Algebra System to draw a graph

**Calculation:**

By using Mathematica,

Use below command:

si=Integrate[Sin[t]/t,{t,0,x}]

Output:

SinIntegral[x]

Use below command:

Plot[si,{x,-4*π,4*π}]

Output:

#### To determine

(b)

**To find:**

Values of x where the function has local maximum values

#### Answer

x=π,-2π, 3π,-4π,5π, -6π

#### Explanation

**Concept:**

i) Fundamental Theorem of Calculus:

If f is a continuous function on [a, b], then the function g defined by

gx=∫axf(t) dt

is continuous on [a, b] and differentiable on (a, b), and g'x=f(x)

ii) Local maxima:

occurs when,

dydx=0 and d2yd2x<0

**Given:**

Six=∫0xsinttdt

**Calculation:**

From the Fundamental theorem,

ft=sintt

Replace t by x,

fx=sinxx

So,S'x=fx=sinxx

Again differentiating S'(x) with respect to x

S''(x)=x·ddxsinx-sinxddxxx2

S''(x)=x·cosx-sinx·1x2

S''(x)=xcosx-sinxx2

By local maxima concept at local maxima

S'x=0

sinxx=0

sinx=0

Taking inverse of sine

x=sin-10

x=(2πn) or x=(2πn+π) We have cos(2πn) =1 and cos(2πn+π)=-1 Hence, at x=2πn+π, S has local maxima when n is positive, and at x=2πn, S has local minima when n is negative

Case 1: For,x=2πn

For n=-1,x=-2π

For n=-2, x=-4π

For n=-3, x=-6π

Case 2: For,x=2πn+π

For n=, x=π

For n=1, x=3π

For n=2, x=5π

For n=3, x=7π

Therefore, x=2πn+π has local maxima when n is positive and x=2πn has local minima when n is negative

To verify the above values, refer the graph of Si(x)

Therefore, local maximum occurs at x=π,-2π, 3π,-4π,5π, -6π,…

#### To determine

(c)

**To find:**

Co-ordinates of the first inflection point to the right of origin

#### Answer

(4.493, 1.6556)

#### Explanation

**Concept:**

i) Fundamental Theorem of Calculus:

If f is a continuous function on [a, b], then the function g defined by

gx=∫axf(t) dt

is continuous on [a, b] and differentiable on (a, b), and g'x=f(x)

ii) Inflection points are the points where f(x) changes from concave upward to concave downward or concave downward to concave upward.

**Given:**

Six=∫0xsinttdt

From part (b),

S'x=fx=sinxx

S''(x)=xcosx-sinxx2

Substitute S''(x)=0,

xcosx-sinxx2=0

xcosx-sinx=0

xcosx=sinx

x=tanx

Draw the graph of y=x that from the origin and y=tan x

The intersection point gives the value of x which is equal to 4.493

To find the y coordinate of the inflection point, evaluate Si4.493

Si4.493=∫04.493sinttdt=1.6556

So, coordinate of the first inflection point to the right of origin are (4.493, 1.6556)

Therefore, the coordinates are (4.493, 1.6556)

#### To determine

(d)

**To check:**

Whether the given function has horizontal asymptotes

#### Answer

Yes

#### Explanation

**Concept:**

i) Fundamental Theorem of Calculus:

If f is a continuous function on [a, b], thenthe function g is defined by

gx=∫axf(t) dt

is continuous on [a, b] and differentiable on (a, b), and g'x=f(x)

ii) Horizontal asymptote:

Line y=L is called a horizontal asymptote of the curve y=f(x) if either

limx→∞f(x)=L or limx→-∞f(x)=L

**Given:**

Six=∫0xsinttdt

**Calculation:**

From part (b),

S'x=fx=sinxx

Now

Use below command

Limit[SinIntegral[x],x→-∞]

Output is:

-π2

Use below command:

Limit[SinIntegral[x],x→∞]

Output:

π2

So, from this f(x) has horizontal asymptotes at y=±π2

Therefore, the given function has horizontal asymptotes.

#### To determine

(e)

**To solve:**

The given equation by using the graph

#### Answer

1.1

#### Explanation

**Given:**

∫0xsinttdt=1

**Calculation:**

First, draw the graph of integral ∫0xsinttdt

Next, draw the graph of the given equation ∫0xsinttdt=1. Both the graphs are shown below.

So, the two functions intersect at point (1.065,1)

Therefore, value of x=1.1 up to two decimals.