Desklib Online Library: Math Quiz #7 Solutions

This article provides step-by-step solutions to Math Quiz #7 problems on area, volume, integrals, power series, and more. Topics covered include finding the area of a region bounded by a graph, evaluating integrals, determining the convergence of improper integrals and series, and using series to evaluate limits.

QUIZ # 7 Problems
(10 questions, 10 points each)
1. Find the area of the region bounded by the graph of 𝑓(𝑥) = 2𝑥2 + 8 and 𝑔(𝑥) =
4𝑥 + 14,at the intervale [−3, 6]
Solution – Area of region bounded by the graph is given by,
∫ |𝑓(𝑥) − 𝑔(𝑥)|𝑑𝑥
𝑏
𝑎
𝑓(𝑥) = 2𝑥2 + 8
𝑔(𝑥) = 4𝑥 + 14
Therefore, Area, A = ∫ |2𝑥2 + 8 − 4𝑥 + 14| 𝑑𝑥
6
−3
Solving the integral, we get A = 290/3 = 96.67
2. Find the volume of the solid generated by rotating the region bounded by 𝑦 = 4𝑥 − 𝑥2
about the line 𝑥 = 5.
Solution –
Given equation, 𝑦 = 4𝑥 − 𝑥2
About the line x = 5 means x = 0 to 5
Volume =
= 130.9
3. Determine if the improper integral converges and, if so, determine its value: ∫ 1
√8−𝑥
3
8
0 𝑑𝑥
Solution –
The problem point is the upper limit
Solving for,
∫ 1
√8−𝑥
3
𝑡
0 𝑑𝑥 = −3(8−𝑡)
2
3
2 + 6
Now Solving for,

lim
𝑡 →8
−3(8−𝑡)
2
3
2 + 6 = 1.61
The limit exists and is finite and so the integral converges and the integral's value is 1.61
4. Evaluate the indefinite integral ∫ 𝑥𝑙𝑛(𝑥)𝑑𝑥
Solution –
Applying Integration by parts,
u = ln x and v = x
Solving, ∫ 𝑥
2 𝑑𝑥 = 𝑥2
4
Therefore, ∫ 𝑥𝑙𝑛(𝑥)𝑑𝑥
5. Evaluate ∫ 4𝑥2+5
√𝑥
3
1 𝑑𝑥
Solution: Applying Integration by parts
u = 4𝑥2 + 5and v = 1
√𝑥
Simplifying,
Substituting in the above equation,
Computing for boundaries

Plugging in x = 1
= 58
5
And,
Substituting all the values,
= 30.66
6. A force of 10 lbs is required to hold a spring stretches 4 inches beyond its natural length.
How much work is done to stretches it from its natural length to 5 inches beyond its
natural length?
Solution –
Force = kx
10 = k (1/3) [4 inches = 1/3 feet]
K = 30
So, f (x) = 30x
Work done W = ∫ 30𝑥
0.416667
0 𝑑𝑥= 30[𝑥2
2 ] = 30 0.4166672
2
= 2.6 ft.lb
7. Integrate ∫ 𝑒3𝑥
𝑒6𝑥+36𝑑𝑥
Solution: ∫ 𝑒3𝑥
𝑒6𝑥+36𝑑𝑥
Consider u = 3x
∫ 𝑒𝑢
3(𝑒2𝑢+36)𝑑𝑢
𝐴𝑝𝑝𝑙𝑦𝑖𝑛𝑔 𝑢 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛, 𝑣 =𝑒𝑢

Applying Integral Substitution, v = 6w
Taking the constant out,
= 1
3 .1
6 tan−1 𝑤
=1
3 .1
6 tan−1 𝑒3𝑥
6
Therefore, ∫ 𝑒3𝑥
𝑒6𝑥+36𝑑𝑥= 1
18 tan−1 1
6 𝑒3𝑥 + C
8. Find the interval of convergence for the power series ∑ 1
2𝑛+2
∞
𝑛=1 𝑥𝑛+2
Solution:
Using the ratio test to find the interval of convergence,
We know that,
Therefore,
Finding,
Simplifying,
Therefore,

Also,
= 1
= |x|
The sum converges for L < 1, therefore solve |x| < 1
For x = 1, there cannot be found a number N which satisfy Cauchy condition, hence it
diverges
For x = -1 = 0 (Converges)
Therefore, the interval of convergence for the power series ∑ 1
2𝑛+2
∞
𝑛=1 𝑥𝑛+2 = -1 ≤ x < 1
9. Does the series ∑ (−1)𝑛 𝑥4𝑛
𝑛!
∞
𝑛=0 converges and, if so to what?
Solution - Using the ratio test to find the interval of convergence,
We know that,
Therefore,
Computing
Solving for,
= −1
𝑛

= 0
L < 1 for every x, therefore ∑ (−1)𝑛 𝑥4𝑛
𝑛!
∞
𝑛=0 Converges for all values of x
10. Use a series to evaluate the limit lim
𝑥→0
sin(𝑥)−𝑥+
1
6𝑥3
𝑥5
Applying L Hospital Rule
Applying L Hospital Rule
Applying L Hospital Rule
Applying L Hospital Rule
= cos 0
120 = 1
120 = 0.0083