Homework 1
Task 1:
Evaluate the following integrals:
1.
´3
−2
x2 − 3 dx
´3
Solution:
2.
´ π/4
0
−2
8
3 −
sec2 θdθ
´ π/4
Solution:
3.
´ π/4
0
8
x2 − 3 dx = 13 x3 − 3x |3−2 = 27
3 − 9 − − 3 − 3 ∗ (−2) =
6
0
π/4
sec2 θdθ = tan θ|0
= tan π4 − tan 0 = 1
tan θ sec θdθ
´ π/4
Solution:
0
π/4
tan θ sec θdθ = sec θ|0
= sec π4 − sec 0 =
√
2−1
Task 2:
Use the Fundamental Theorem of Calculus to nd the derivative of the
functions dened below
1. F (x) =
´x
1
dt
1 t3 +1
Solution:
0
´
0
´
The FTC states that if (F (x)) = f (x), then f (x) dx =
0
F (x)+C . If we take the derivative on both sides, then F (x) =
1
x3 +1
2. F (x) =
´x√
Solution:
0
t2 + 4dt
The FTC states that if (F (x)) = f (x), then f (x) dx =
0
F (x)+C . If we take the derivative on both sides, then F (x) =
√
t2 + 4
Task 3:
Evaluate the integral by making use of the given substitution
1
1.
´
√
x2 x3 + 1dx, u = x3 + 1
Solution:
2.
´
By
the given ´substitution,
we see that du = 3x2 dx. So
√
´ 2using
1√
2 3/2
3
x x + 1dx = 3 udu = 9 u
+ C . By substituting
3/2
´ 2√
2
3
3
+ C.
x + 1 = u, we get x x + 1dx = 9 x3 + 1
cos3 θ (sin θ) dθ, u = cos θ
Solution:
By ´using the given substitution,
we see that du = sin θdθ.
´
So cos3 θ (sin θ)´dθ = u3 du = 14 u4 + C . By substituting
sin θ = u, we get cos3 θ (sin θ) dθ = 14 cos4 θ + C ,
Task 4:
Evaluate the indenite integral
1.
´
esin x (cos x) dx
Solution:
2.
´
√
sin
√ x dx
x
Solution:
Make
´ sin xthe substitution
´ u = sin x. Then du = cos xdx. So
e
(cos x) dx = eu du = eu + C = esin x + C .
[Hint: use u =
√
x]
Make the substitution u =
´
√
sin
√ x dx
x
´
√
x. Then du =
1 √1
2 x dx.
√
So
= 2 sin udu = −2 cos u + C = −2 cos x + C .
Task 5:
Evaluate the integral
1.
´
x cos (5x) dx
Solution:
2.
Let f (x) = x ´and g (x) = cos (5x). Then by ´using integration by parts, x cos (5x) dx = 51 x sin (5x) − 51 sin (5x) dx =
1
1
5 x sin (5x) + 25 cos (5x) + C .
0
´
(ln x) dx [Hint: You can think of 1 as g (x) in the formula
´ 0
f (x) g (x) − f (x) g (x) dx]
Solution:
0
´
0
f (x) g (x) dx =
Let f (x)
´ = ln x and g (x) =´ 1. Then by using integration by
parts, (ln x) dx = x ln x − dx = x ln x − x + C .
0
2