Ali Alderson, Nick Birkby, Kenzie Rairdon
5/13/14
Chapter 7 Practice Test
1. Set up the integral to find the area bounded by y2 = 4 – x and x + 2y – 1 = 0
1. a) Graph the equations
b) Put the two equations in x = f (y) form
y2 = 4 – x becomes x = 4 – y2
x + 2y – 1 = 0 becomes x = -2y + 1
c) Find the y-values of the points of intersection
4 – y2 – (-2y + 1) = 0
-y2 + 2y + 3 = 0
y2 – 2y – 3 = (y+1) (y-3) = 0
y = -1, 3
d) Find the integral
3
∫−1(−𝑦 + 2𝑦 + 3)𝑑𝑦
2. Find the area bounded by y = - x2 and y = x2 + 8
a) Graph the equations
b) Find the x-values of the points of intersection by setting the two equations equal to
one another
(- x2) – (x2 – 8) = 0
- 2x2 + 8 = 0
x2 – 4 = (x + 2) (x – 2) = 0
x = - 2, 2
c) Set up an integral with limits equal to the points from step b
2
∫−2(−2𝑥 2 + 8)𝑑𝑥
d) Solve the integral
− 𝟐𝒙𝟑
𝟑
−2
+ 8𝑥 from – 2 to 2
−2
[ 3 (2)3 + 8 (2)] – [
64
or 21.33
3
3
(−2)3 + 8 (−2)]
3. Find the area bounded by y = x2 + 1 and y = 5
a) Graph the equations
b) Find the x-values of the points of intersection
5 – (x2 + 1) = 0
-x2 + 4 = 0
x2 – 4 = (x + 2) (x – 2) = 0
x = -2, 2
c) Solve the integral
2
∫ (−𝑥 2 + 4)𝑑𝑥
−2
−(𝑥)3
+ 4𝑥 from − 2 to 2
3
[
−(2)3
3
32
3
or 10.667
4. Find the area analytically
2
2 (∫0 2x^2 – x^4 + 2x^2) dx
4x^2- x^4
2 (4x^3/ 3 – x^5/5 ) |02
2 (32/3- 32/5)
2 ((32 * 5 / 3*5)- (32 *3/ 5 * 3))
2(160/15- 96/15)
2(64/15)
= 128/15
3
+ 4(2)] - [−(−2)
3
+ 4(−2)]
5.
Find the area of the regions enclosed by the lines and curves
Use graphing calculator to graph
x= -y^2
x= -3y^2 +2
1
∫0 2 – 2y^2 dy
A= 2
2 (2y – 2y^3/3) |01
2 (6/3 – 2/3)
2/1 * 4/3
= 8/3
6.
Find the area of the regions enclosed by the lines and curves
Graph on a graphing calculator
𝜋/2
A= ∫0
3sin(y) (cos(y))^1/2 dy
u= cos(y)
du= -sin(y) dy
-du= sin(y) dy
-3du (u)^1/2
-3 (2/3 u ^3/2)
-3 * 2/3= -2
-2 ( cos 3/2) |0π/2
-1 * -2= 2
7.
Find the area of the regions enlaces by the lines and curves
1
∫−1(7 − 2x^2) – (x^2 + 4) dx
1
∫−1 3 − 3x^2 dx
3x- x^3 |-11
(3- 1) – (-3 +1)
2 – (- 2)
=4
8.
True or False: If a function y = f(x) is continuous on an interval [a,b], then the length of its curve
2
𝑏
𝑑𝑦
is given by ∫𝑎 √1 +
𝑑𝑥 . Justify your answer
𝑑𝑥
False.The Function must also be differentiable on the interval.
9.
𝑥
Find the length of the curve ∫0 √cos 2𝑡 dt from x=0 x=𝜋/4 (21)
10.Find a curve through the point (0,1) whose length integral is
2
L= ∫1 √1 +
1
𝑦4
dy
In simpler terms, Take the anti-derivative to get the general equation and use the points
to solve for c!
Then, find how many such curves there are. Give reasons for your answer.
There is only 1 equation due to the given point of (0,1)
For questions 11-13, find the volume of the solid generated by revolving the region bounded by the lines
and curves about the x-axis.
11.
Y=𝑥 2 , y=0, x=2
12.Y=sec x, y=√2, -𝜋4 ≤x≤ 𝜋/4
13. Y= 𝑥 3,y=0, x=2