Calculus Quiz 7: Definite Integrals, Series Convergence, and More

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Added on  2023/06/14

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This document presents solutions to a Calculus Quiz (Quiz 7) consisting of 10 problems. The problems cover a range of topics including finding the area of a region bounded by graphs, calculating the volume of a solid generated by rotating a region around a line, determining the convergence of improper integrals, evaluating indefinite integrals, applying integration by parts, calculating work done by a spring, integrating complex functions using substitution, finding the interval of convergence for power series, determining series convergence, and evaluating limits using series and L'HΓ΄pital's Rule. Each solution provides step-by-step calculations and explanations to arrive at the final answer. Desklib provides more solved assignments and study tools for students.
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QUIZ # 7 Problems
(10 questions, 10 points each)
1. Find the area of the region bounded by the graph of 𝑓(π‘₯) = 2π‘₯2 + 8 and 𝑔(π‘₯) =
4π‘₯ + 14,at the intervale [βˆ’3, 6]
Solution – Area of region bounded by the graph is given by,
∫ |𝑓(π‘₯) βˆ’ 𝑔(π‘₯)|𝑑π‘₯
𝑏
π‘Ž
𝑓(π‘₯) = 2π‘₯2 + 8
𝑔(π‘₯) = 4π‘₯ + 14
Therefore, Area, A = ∫ |2π‘₯2 + 8 βˆ’ 4π‘₯ + 14| 𝑑π‘₯
6
βˆ’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 = 4π‘₯ βˆ’ π‘₯2
about the line π‘₯ = 5.
Solution –
Given equation, 𝑦 = 4π‘₯ βˆ’ π‘₯2
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
√8βˆ’π‘₯
3
8
0 𝑑π‘₯
Solution –
The problem point is the upper limit
Solving for,
∫ 1
√8βˆ’π‘₯
3
𝑑
0 𝑑π‘₯ = βˆ’3(8βˆ’π‘‘)
2
3
2 + 6
Now Solving for,
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lim
𝑑 β†’8
βˆ’3(8βˆ’π‘‘)
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 ∫ π‘₯𝑙𝑛(π‘₯)𝑑π‘₯
Solution –
Applying Integration by parts,
u = ln x and v = x
Solving, ∫ π‘₯
2 𝑑π‘₯ = π‘₯2
4
Therefore, ∫ π‘₯𝑙𝑛(π‘₯)𝑑π‘₯
5. Evaluate ∫ 4π‘₯2+5
√π‘₯
3
1 𝑑π‘₯
Solution: Applying Integration by parts
u = 4π‘₯2 + 5and v = 1
√π‘₯
Simplifying,
Substituting in the above equation,
Computing for boundaries
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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 = ∫ 30π‘₯
0.416667
0 𝑑π‘₯= 30[π‘₯2
2 ] = 30 0.4166672
2
= 2.6 ft.lb
7. Integrate ∫ 𝑒3π‘₯
𝑒6π‘₯+36𝑑π‘₯
Solution: ∫ 𝑒3π‘₯
𝑒6π‘₯+36𝑑π‘₯
Consider u = 3x
∫ 𝑒𝑒
3(𝑒2𝑒+36)𝑑𝑒
𝐴𝑝𝑝𝑙𝑦𝑖𝑛𝑔 𝑒 π‘ π‘’π‘π‘ π‘‘π‘–π‘‘π‘’π‘‘π‘–π‘œπ‘›, 𝑣 =𝑒𝑒
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Applying Integral Substitution, v = 6w
Taking the constant out,
= 1
3 .1
6 tanβˆ’1 𝑀
=1
3 .1
6 tanβˆ’1 𝑒3π‘₯
6
Therefore, ∫ 𝑒3π‘₯
𝑒6π‘₯+36𝑑π‘₯= 1
18 tanβˆ’1 1
6 𝑒3π‘₯ + C
8. Find the interval of convergence for the power series βˆ‘ 1
2𝑛+2
∞
𝑛=1 π‘₯𝑛+2
Solution:
Using the ratio test to find the interval of convergence,
We know that,
Therefore,
Finding,
Simplifying,
Therefore,
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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 βˆ‘ 1
2𝑛+2
∞
𝑛=1 π‘₯𝑛+2 = -1 ≀ x < 1
9. Does the series βˆ‘ (βˆ’1)𝑛 π‘₯4𝑛
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,
= βˆ’1
𝑛
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= 0
L < 1 for every x, therefore βˆ‘ (βˆ’1)𝑛 π‘₯4𝑛
𝑛!
∞
𝑛=0 Converges for all values of x
10. Use a series to evaluate the limit lim
π‘₯β†’0
sin(π‘₯)βˆ’π‘₯+
1
6π‘₯3
π‘₯5
Applying L Hospital Rule
Applying L Hospital Rule
Applying L Hospital Rule
Applying L Hospital Rule
= cos 0
120 = 1
120 = 0.0083
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