Download Solutions to Midterm 1 and more Exams Calculus in PDF only on Docsity!
Solutions to Midterm 1
- §5.2 #32: The graph of g consists of two straight lines and a semicircle. Use it to evaluate each integral.
(a)
0
g(x) dx Solution: It’s a triangle with base = 2 and height = 4, so the area is 4. (b)
2
g(x) dx Solution: It’s a semi-circle with radius = 2; the area of the whole circle would be π · 22 = 4π, so the area of the semi-circle is 2π. But it’s below the x-axis, so the integral is − 2 π.
(c)
0
g(x) dx Solution: It’s the sum previous two, plus a triangle with area 1 /2, so the integral is 4 − 2 π + 1/2.
- §5.3 #7: Evaluate (^) ∫ 0 − 1
(2x − ex) dx.
Solution: ∫ (^0)
− 1
(2x − ex) dx =
[
x^2 − ex
] 0
− 1 = (0 − 1) − (1 − e−^1 ) =^1 e
- §5.5 #44: Evaluate (^) ∫ √ π 0
x cos(x^2 ) dx.
Solution: Let u = x^2 , so du = 2x dx, so x dx = 12 du. Then ∫ (^) x=√π
x=
x cos(x^2 ) dx =
∫ (^) u=π
u=
cos u · 12 du
= 12 sin u
π
0 = 12 · 0 − 12 · 0 = 0.
- §5.5 #55: Evaluate (^) ∫ e^4 e
dx x
ln x
Hint: Substitute u = ln x. Solution: Let u = ln x, so du = 1/x dx. Then ∫ (^) x=e 4
x=e
dx x
ln x
∫ (^) u=
u=
√^ du u
=
1
u−^1 /^2 du
= 2u^1 /^2
4
1 = 2 · 2 − 2 · 1 = 2.
Then we let u = sin x, so du = cos x dx, so this becomes ∫ √ 2 / 2
1
u^5 (1 − u^2 ) du =
1
(u^5 − u^7 ) du
=
[ (^) u 6 6
− u
8 8
]√ 2 / 2
1
- §5.7 #21: Use partial fractions to evaluate ∫ 5 x + 1 (2x + 1)(x − 1)
dx.
Solution: We wish to find numbers A and B such that 5 x + 1 (2x + 1)(x − 1) =^
A
2 x + 1 +^
B
x − 1
Ax − A + 2Bx + B (2x + 1)(x − 1). Thus we should solve
A + 2B = 5 −A + B = 1.
Adding the two lines we get 3B = 6, so B = 2, and thus A = 1. Thus ∫ (^5) x + 1 (2x + 1)(x − 1) dx^ =
∫ (^1
2 x + 1 +^
x − 1
dx
=
2 ln^ |^2 x^ + 1|^ + 2 ln^ |x^ −^1 |^ +^ C.
- §5.8 #9: Evaluate (^) ∫ (^) ∞
4
e−y/^2 dy.
Solution: On the one hand, we could say ∫ (^) ∞
4
e−y/^2 dy = − 2 e−y/^2
∞ 4 = 0 − − 2 e−^2
=
e^2. If you’re not comfortable finding an anti-derivative for e−y/^2 in one step, you could substitute u = −y/2, so dy = −dy/2, so dy = − 2 du: ∫ (^) ∞
4
e−y/^2 dy =
− 2
eu(− 2 du)
−∞
2 eudu
= 2e−^2 − 0 =^2 e^2
