# Integral problems

1. Dec 16, 2007

### phantomprime

1. The problem statement, all variables and given/known data
1)The graph of g consists of two straight lines and a semicircle Use it to evaluate each integral

a)
b)
c)

Note: graph semi circle r= 2 it goes from 2,-6 I couldn't scan the page I have

2. Relevant equations

3. The attempt at a solution

a) y= mx+b
(-2x+y)dx|2/0
-2((2)^2)/2 + 4(2) = 4

b) sqrt ((-x^2)-8x-12)dx|6/2
u=((-x^2)-8x-12)
du=((-x^3)/3)-(4x^2)-12x)^2/3

2/3 (U^2/3)

after all the calculation i get 2 Sqrt 24 - 4 Sqrt 10
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2) PG358 #58
Water flows from the bottom of a storage tank at a rate of r(t) = 200-4t liters per minute, where 0 <\ t <\50. Find the amount of water that flow from the tank during the first 10
minutes.

attempt (top of page)..(yes i know I have terrible handwriting)

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3) Pg366 #60
The rate of production of calculators are in "t" weeks is
dx/dt = 5000(1- (100/(t+10)^2)) cal/week

The production approaches 5000per week as time goes on, the initial production is lower because workers are not familiar with new techniques. Find the number of calculators from the beginning of the third week to the end of the fourth week.

attempt (bottom of page)

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find the volume of the solid obtained by rotating about the y-axis region bounded by the curves y= e^ (-x^2), y=0 x=0 x=1

attempt:

Last edited: Dec 16, 2007
2. Dec 16, 2007

### HallsofIvy

Staff Emeritus
Wouldn't it be easier to just use the fact that the integral is the area under the curve?
$\int_0^2 g(x)dx$ is just the area of a triangle.
$\int_2^6 g(x)dx$ is the negative of the area of a semicircle.
$\int_0^7 g(x)dx$ is the sum of the two areas in (1) and (2) plus the area of another triangle.

First some technical points- when you wrote the integral, you didn't include "dt". That's important. Second, after you have integrated do not include the integral sign again- the standard notation is a vertical line after the anti-derivative with the lower and upper bounds at bottom and top. The calculation, however, looks corrrect.

The integral should not be from 3 to 4. The problem says "from the beginning of the third week to the end of the fourth week". If it had said "from the beginning of the third week to the end of the fourth week", would you have used 3 as both lower limit and upper limit? Also the integral of $100/u^2$ is not $100/(u^3/3)$. The "power rule" for integration is (1/(n+1)u^(n+1)) and here, n= -2. Finally, when you change the variable by substitution, change the limits of integration. If u= t+ 1, when t= 3, u= 3+ 1= 4 and when t= 5, u= 5+1= 6.