Q1 :- Use integration by substitution to solve the integral below.

\[\int { { _{ 4e }{ -7x }_{ dx } } }\] u=-7x du=-7dx dx=dU/-7 \[4\int { { e }^{ u } } \frac { du }{ -7 }\] \[\frac { 4 }{ -7 } \int { { e }^{ u } } du\] \[\frac { 4 }{ -7 } { e }^{ u }+\quad Constant\] as u= -7x

\[\frac { 4 }{ -7 } { e }^{ -7x }+\quad Constant\]

Q2:-Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of dN/dt = -3150/t^4 - 220 bacteria per cubic centimeter per day, where t is the number of days since treatment began. Find a function N( t ) to estimate the level of contamination if the level after 1 day was 6530 bacteria per cubic centimeter.

N’(t)=(???3150/ t4)-220 \[N(t)\quad =\quad \int { (\frac { -3150 }{ { t }^{ 4 } } -220)dt } \]

when N=1, then

N(1)=1050 - 220 + Constant=6530, hence Constant=5700

The function is

N(t) = 1050 / t3 -220t +5700

Q3:- Find the total area of the red rectangles in the figure below, where the equation of the line is f(x) = 2x - 9.

A : - Each square in the graph has an area of 1. Each rectangle has a width of 1. Counting each rectangle left to right the areas are

Area=1+3+5+7=16.

But an better way would be to use integral to find the area

findArea <- function(x)
  {
  2*x-9
}

integrate(findArea, lower = 4.5, upper = 8.5)
## 16 with absolute error < 1.8e-13

Q4 :- Find the area of the region bounded by the graphs of the given equations.

     y=x^2???2x???2   
     y=x+2

A:-

curve(x^2 - 2*x - 2, -5, 5)
curve(x + 2, -5, 5, add=T, col="red")

((3/2)*4^2 +4*4 -(1/3)*4^3) - ((3/2)*(-1)^2 +4*(-1) -(1/3)*(-1)^3)
## [1] 20.83333

The formula for finding the area enclosed by two curves is as follows :-

\[\int _{ a }^{ b }{ (top-bottom)dx } \] Where a = -1 & b= 4 (We should get two values for x because of the quadratic term. x^2 -2x-2 = x+2 ~ (x-4)(x+10 )

Function :-

\[\int _{ -1 }^{ 4 }{ (x+2)-({ x }^{ 2 }-2x-2)dx } \] \[\int _{ -1 }^{ 4 }{ -{ x }^{ 2 }+3x+4)dx }\] Using R in the above function(x)

findArea <- function(x)
  {
  -x^{2}+3*x+4
  }


## integrate the function from 0 to infinity
integrate(findArea, lower = -1, upper = 4)
## 20.83333 with absolute error < 2.3e-13

Q5 :- A beauty supply store expects to sell 110 flat irons during the next year. It costs $3.75 to store one flat iron for one year. There is a fixed cost of $8.25 for each order. Find the lot size and the number of orders per year that will minimize inventory costs.

A:-

Assume Y = cost n = the no of orders per year x = no of irons in order. thus nx=110 so x=110n, assume half an order is in storage at on average. Such that,

\[Y\quad =8.25n\quad +\quad \frac { 3.75x }{ 2 } \] if nx = 100 , then x= 100/n

\[Y\quad =8.25n\quad +\quad \frac { 206.25 }{ n } \]

\[Y`\quad =8.25\quad +\quad \frac { 206.25 }{ n^2 } \]

if C`=0, then

\[n\quad =\quad \sqrt { \frac { 206.25 }{ 8.25 } } \] thus solving the above , we get n = 5 orders per year which will keep the inventory cost to minimum

Q6 :- Use integration by parts to solve the integral below.

   $$ \int { ln(9x).{ x }^{ 6 }dx }  $$

A :-

\[ \int { udv\quad =\quad uv\quad -\quad \int { vdu } } \] u=ln(9x) dv=x^6dx

\[\frac { 1 }{ 7 } ln(9x){ x }^{ 7 }\quad -\quad \frac { 1 }{ 7 } \int { { x }^{ 6 }dx } \quad \]

\[ \frac { 1 }{ 7 } ln(9x){ x }^{ 7 }\quad -\quad \frac { 1 }{ 7 } (\frac { { x }^{ 7 } }{ 7 } )\quad +\quad C\quad \quad \] \[ \frac { { x }^{ 7 } }{ 7 } \left[ n(9x)\quad -\frac { 1 }{ 7 } \right] \quad +\quad C\quad \quad \] ## Q 7 :- Determine whether f ( x ) is a probability density function on the interval 1, e6 . If not, determine the value of the definite integral. f( x ) = 1 / 6x.

A:-

\[ \int _{ 1 }^{ { e }^{ 6 } }{ \frac { 1 }{ 6x } } dx\quad \quad \] \[ \frac { 1 }{ 6 } \int _{ 1 }^{ { e }^{ 6 } }{ \frac { 1 }{ x } } dx\quad \quad\]

\[ \frac { 1 }{ 6 } (ln({ e }^{ 6 })\quad -ln(1))\quad \] \[\frac { 1 }{ 6 } (6-0)\quad =\quad 1\] Hence the definite integral of the function on interval [1,e6] is 1