Cal 2 Midterm #1 Questions

Question 1

What is this integral in Trigonometric Form ?

Answer 1

We can set the sides of the triangle to:

O = x; A = 2; H = √(4 + x²)

Now we can say:

√(4 + x²) = 2secθ
x = 2tanθ
dx = 2sec²θ dθ

Then fill in the values:

∫ √(4 + x²)dx = ∫ 2secθ⋅2sec²θ dθ = ∫ 4sec³θ dθ

Question 1

Explain how to solve:

Answer 1

Since we have the value of ∫ sec²x[ƒ(x)]dx = - π/4,

We can assume that this is in the form of : ∫ (d𝓾)(𝓿),
where 𝓿 = ƒ(x); and d𝓾 = sec²x

So we can do integration by parts using this values:

𝓾 = tanx; d𝓾 = sec²x dx
𝓿 = ƒ(x); d𝓿 = ƒ’(x) dx

Now solve at the bounds:

∫ tanx[ƒ(x)]dx = (𝓾)(𝓿) - ∫ 𝓿d𝓾

[tanx][ƒ(x)] - ∫ sec²x[ƒ(x)]dx

[tanx][ƒ(x)] + π/4

Question 2

Explain solution to:

Answer 2

Start with knowing that ∫ sin²t = ½t - ½[sin(t)cos(t)]

Then evaluate at π/4:

½[π/4] - ½[sin(π/4)cos(π/4)] → [π/8] - ½[√2/2 * √2/2] → [π/8] - ½[2/4]

(π - 2)/8

Question 3

Explain how to solve which of these options matches the Trig values of:

x -3 = 4sinθ

Answer 3

We have to adjust each integral to find the triangle which would satisfy the given Trig Value:

√(7 + 6x - x²); We can use completing the square to solve this:

√-(x² - 6x + 9 - 9 - 7) → - [(x - 3)² - 16] → √[16 - (x - 3)²]

O = (x - 3); A = √[16 - (x - 3)²]; H = 4

so sinθ = (x - 3)/4 → 4sinθ = (x - 3)

Question 4

What is the first step in solving this integral

Answer 4

Since the degree in the numerator is one above that in the denominator we should do Polynomial Long Division

Question 5

How to solve this integral

Answer 5

Between these bounds, this function is divergent

Question 6

How is this improper integral solved ?

Answer 6

Question 7

What is the formula for Surface Area of a function revolved around an axis ?

Answer 7

Question 8

What is the largest set of values for r such that the sequence:

limit n→∞ rⁿ

Answer 8

when are is -1 < r ≤ 1

since -1∞ oscillates between + and -

Any fraction to the ∞ power will tend to 0

And 1∞ is the last value that doesnt tend to ∞

Question 9

How to set up this integral

Answer 9

A/x + B/x² + (Cx+D)/(x²+1)

Question 10

What is first step to setting up this integral

Answer 10

Start by separating cos³x into cos²xcosx then turning cos²x into [1- sin²x]

∫ sin³/²x[1- sin²x]cosx dx

Now we can set 𝓾 = sinx; d𝓾 = cosx, then solve the problem out

Question 11

What is the partial fraction decomposition of:

Answer 11

Question 12

Explain the solution to this problem:

Answer 12

We start by taking the limit of the sequence as it approaches infinity and instantly recognize that the (-1) in the numerator of the right side makes the sequence Non-Monotonus since the negative forces some oculation of the function.

The second is that since the left side. as n approaches ∞ is linear we know that this function is also boundless

Question 13

Explain Convergence/Divergence of this function:

Answer 13

We can start by creating a chart of the first 5 terms in the series.

Because sin(nπ/2) has the nature to oscillate between +/-, we know that the function is Non-Monotonus

Second, we notice the value of the function tends to shrink as we have a growth of n. This must also indicate convergence.

Lastly we can do Squeeze Theorem given that our function sin(nπ/2) never exceeds -1 or 1, and taking the limit as n approaches ∞

-1/n < sin(nπ/2) < 1/n = 0 ; Convergent

Question 14

Explain the solution to this problem

Answer 14

We start by going ahead and setting a limit as n approaches ∞

We now that tan⁻¹(∞) = π/2

We also know that n = ∞

Thus our function is ∞ ⋅ π/2 which is monotonically increasing however boundlessly since it is being increased arbitrarily

Question 15

How to begin this integral ?

Answer 15

Notice that sin³x can be turned into a [1-cos²x] while leaving a sinx available to use in u-substitution