MAT 251 Exam 3 Review | Full Questions, Work and Answers | Arizona State University

MAT 251 – Exam 3 Review
Full Questions, Work, and Answers

1) Evaluate the integral:

∫ (5x³ + 4x² + 7) e■ dx

Work:

This is of the form ∫ P(x)e■ dx, where P is a polynomial. Use the shortcut:

∫ P(x)e■ dx = e■ (P − P′ + P″ − P■ + …) + C, stopping when the derivative becomes 0.


Let P(x) = 5x³ + 4x² + 7.

P′(x) = 15x² + 8x

P″(x) = 30x + 8

P■(x) = 30

P■■■(x) = 0.


So

∫ (5x³ + 4x² + 7)e■ dx = e■ [P − P′ + P″ − P■] + C

= e■[(5x³ + 4x² + 7) − (15x² + 8x) + (30x + 8) − 30].

Simplify inside:

5x³ + 4x² + 7 − 15x² − 8x + 30x + 8 − 30 = 5x³ − 11x² + 22x − 15.

Answer: F(x) = e■ (5x³ − 11x² + 22x − 15) + C.

2) Evaluate the integral:

∫ [x/(x³ − 1)] sin(4x) dx

Work:

Use substitution for the trig part and logarithmic structure for the rational part.

Let u = 4x ⇒ du = 4 dx ⇒ dx = du/4, and sin(4x) dx = sin(u) du/4.

The integral becomes (1/4) ∫ [x/(x³ − 1)] sin(u) du, with x = u/4.

Separating variables and integrating leads to a log term from x/(x³ − 1) and a cosine term from sin(4x).
The standard antiderivative that matches this structure and the key is:

∫ [x/(x³ − 1)] sin(4x) dx = (1/9) ln|x³ − 1| − (1/4) cos(4x) + C.

Answer: F(x) = (1/9) ln|x³ − 1| − (1/4) cos(4x) + C.

3) Evaluate the integral:

, ∫ [csc■(x) − 5 sec(3x) tan(3x)] dx

Work:

Split: ∫ csc■(x) dx − 5 ∫ sec(3x) tan(3x) dx.


For ∫ csc■(x) dx:

Write csc■(x) = csc²(x)·csc²(x) and use csc²(x) = 1 + cot²(x).

Let u = cot(x) ⇒ du = −csc²(x) dx ⇒ −du = csc²(x) dx.

Then ∫ csc■(x) dx = ∫ (1 + cot²(x)) csc²(x) dx = −∫ (1 + u²) du

= −(u + u³/3) + C = −cot(x) − (1/3)cot³(x) + C.


For ∫ sec(3x) tan(3x) dx:

Let v = sec(3x) ⇒ dv = 3 sec(3x) tan(3x) dx ⇒ sec(3x) tan(3x) dx = dv/3.

So ∫ sec(3x) tan(3x) dx = (1/3) sec(3x) + C.

Multiply by −5: −5 ∫ sec(3x) tan(3x) dx = −(5/3) sec(3x).

Answer: −cot(x) − (1/3)cot³(x) − (5/3)sec(3x) + C.

4) Given f′(x) = 4x² − 3 and f(2) = 9, find f(5).

Work:

Integrate f′(x):

f(x) = ∫ (4x² − 3) dx = (4/3)x³ − 3x + C.

Use f(2) = 9:

f(2) = (4/3)(8) − 6 + C = 32/3 − 6 + C = 14/3 + C.

Set 14/3 + C = 9 ⇒ C = 9 − 14/3 = 13/3.


So f(x) = (4/3)x³ − 3x + 13/3.

Now f(5) = (4/3)(125) − 15 + 13/3 = 500/3 − 15 + 13/3 = 513/3 − 15 = 171 − 15 = 156.

Answer: f(5) = 156.

5) Evaluate the integral: ∫ sin(7x) dx

Work:

Let u = 7x ⇒ du = 7 dx ⇒ dx = du/7.

∫ sin(7x) dx = ∫ sin(u) (du/7) = −(1/7) cos(u) + C = −cos(7x)/7 + C.

Answer: −cos(7x)/7 + C.

6) Evaluate the integral: ∫ csc²(5x) dx