The Indefinite Integral
Solutions To Selected Problems
Calculus 9th Edition Anton, Bivens, Davis
Matthew Staley
November 15, 2011
1. Confirm that the formula is correct, and state a corresponding integration formula.
(a)
d √
x
[ 1 + x2 ] = √
dx
1 + x2
1
d √
[ 1 + x2 ] = (1 + x2 )−1/2 (2x)
dx
2
2x
= √
2 1 + x2
x
=√
1 + x2
Z
√
x
√
dx = 1 + x2 + C
1 + x2
(b)
d 1
3
sin(1 + x ) = x2 cos(1 + x3 )
dx 3
1
d 1
3
sin(1 + x ) = cos(1 + x3 )(3x2 )
dx 3
3
= x2 cos(1 + x3 )
Z
1
x2 cos(1 + x3 ) dx = sin(1 + x3 ) + C
3
2. Find the Derivative and state a corresponding integration formula.
(a)
d √ 3
1
[ x + 5] = (x3 + 5)−1/2 (3x2 )
dx
2
3x2
= √
2 x3 + 5
Z
√
3x2
√
dx = x3 + 5 + C
2 x3 + 5
1
(b)
√
√ d √
d
[sin(2 x)] = cos(2 x) (2 x)
dx
dx
√
1
= cos(2 x)(2 √ )
2 x
√
cos(2 x)
√
=
x
√
Z
√
cos(2 x)
√
dx = sin(2 x) + C
x
3. Evaluate the integral by rewriting the integrand appropriately, if required, and
applying the power rule.
(a)
(b)
Z
Z
x8 dx =
x5/7 dx =
Z
(c)
x8+1
x9
+C =
+C
8+1
9
x
3
√
x5/7+1
x12/7
7 12/7
+ C = 12 + C =
x
+C
5/7 + 1
12
7
Z
x3 x1/2 dx
x dx =
Z
x3+1/2 dx
=
Z
x7/2 dx
=
x7/2+1
+C
7/2 + 1
x9/2
= 9 +C
=
2
=
2 9/2
x +C
9
2
4. Evaluate each integral by applying Theorem 4.2.4 and formula 2 in Table 4.2.1
appropriately,
(a)
Z 2
5x + 5 dx
3x
Z
Z
2
= 5 x dx +
x−5 dx
3
2 x−4
1 2
=5 x +
+C
2
3 −4
=
1
5 2
x − 4 +C
2
6x
Z
(b)
−3
x − 3x1/4 + 8x2 dx
Z
=
=
x
−3
Z
dx − 3
x
1/4
Z
dx + 8
x2 dx
x1/4+1
x3
x−2
−3
+8 +C
−2
1/4 + 1
3
1
= − 2 −3
2x
= −
4
8
x5/4 + x3 + C
5
3
1
12
8
+ x5/4 + x3 + C
2
2x
5
3
3
5. Evaluate each integral.
Z
(a)
x(1 + x2 ) dx
Z
(x + x4 ) dx
Z
Z
=
x dx + x4 dx
=
=
x2 x5
+
+C
2
5
Z
(b)
Z
=
x1/3 (2 − x)2 dx
x1/3 (4 − 4x + x2 ) dx
Z
4x1/3 − 4xx1/3 + x2 x1/3 dx
Z
Z
Z
1/3
4/3
= 4 x dx − 4 x dx + x7/3 dx
=
x1/3+1
x4/3 + 1
x7/3+1
−4
+
+C
1/3 + 1
4/3 + 1
7/3 + 1
3
3
3
= 4 x4/3 − 4 x7/3 + x10/3 + C
4
7
10
3
12
= 3x4/3 − x7/3 + x10/3 + C
7
10
=4
x5 + 2x2 − 1
dx
x4
Z 5
Z
Z
Z
x
2x2
1
−2
=
+ 4 − 4 dx =
x dx + 2 x dx − x−4 dx
4
x
x
x
x2 2x−1 x−3
=
+
−
+C
2
−1
−3
x2 2
1
=
− + 3 +C
2
x 3x
Z
(c)
4
(d)
Z
(3 sin(x) − 2 sec2 (x)) dx
Z
Z
= 3 sin(x) dx − 2 sec2 (x) dx
= −3 cos(x) − 2 tan(x) + C
Z
(e)
sec(x)(sec(x) + tan(x)) dx
Z
(sec2 (x) + sec(x) tan(x)) dx
Z
Z
2
sec (x) dx + sec(x) tan(x) dx
=
=
= tan(x) + sec(x) + C
Z
(f)
sec(θ)
dθ
cos(θ)
Z
=
sec(θ) sec(θ) dθ
Z
=
(g)
(h)
sec2 (θ) dθ = tan(θ) + C
Z
sin(x)
dx
cos2 (x)
Z
sin(x) 1
=
dx
cos(x) cos(x)
Z
=
tan(x) sec(x) dx = sec(x) + C
Z
Z
1
[1 + sin (y) csc(y)] dy =
1 + sin2 (y)
dy
sin(y)
Z
Z
Z
= [1 + sin(y) ]dy =
dy + sin(y) dy = y − cos(y) + C
2
5
Z
(i)
1
dx
1 + sin(x)
Z
1
1 − sin(x)
·
dx
1 + sin(x) 1 − sin(x)
Z
1 − sin(x)
dx
1 − sin2 (x)
Z
1 − sin(x)
dx
cos2 (x)
Z
sin(x)
1
−
dx
2
cos (x) cos(x) cos(x)
=
=
=
=
Z
=
Z
2
sec (x) dx −
tan(x) sec(x) dx
= tan(x) − sec(x) + C
6. A particle moves along an s-axis with position function s = s(t) and velocity
function v(t) = s0 (t). Use the given information to find s(t).
(a) v(t) = 32t;
s(0) = 20
Z
s(t) =
0
Z
s (t) dt =
Z
=
32t dt = 32
v(t) dt
t2
+ C = 16t2 + C
2
s(0) = 16(0)2 + C = 20
C = 20
s(t) = 16t2 + 20
6
√
(b) v(t) = 3 t;
s(4) = 1
Z
s(t) =
v(t) dt
Z
=
2
3t1/2 dt = 3 t3/2 + C = 2t3/2 + C
3
s(4) = 2(4)3/2 + C = 1
16 + C = 1
C = −15
s(t) = 2t3/2 − 15
7. Find an equation of the curve that satisfies the given conditions.
(a) At each point (x, y) on the curve the slope is 2x + 1; the curve passes
through the point (−3, 0).
What we are given is that the derivative of the curve is 2x + 1, so we
can find the curve by integrating:
Z
2x + 1 dx = 2
x2
+ x + C = x2 + x + C = f (x)
2
f (−3) = (−3)2 − 3 + C = 0
6+C =0
C = −6
f (x) = x2 + x − 6
7
(b) At each point (x, y) on the curve the slope is − sin(x); the curve passes
through the point (0, 2)
Z
f (x) =
− sin(x) dx = cos(x) + C
f (0) = cos(0) + C = 2
1+C =2
C=1
f (x) = cos(x) + 1
(c) At each point (x, y) on the curve, y satisfies the condition d2 y/dx2 = 6x;
the line y = 5 − 3x is tangent to the curve at the point where x = 1.
First find
dy
dx
=
R
d2 y
dx2
dx =
R
6x dx = 3x2 + C.
At the point x = 1, the slope of the tangent line there is −3, hence
3(1)2 + C = −3 → C = −6
So
dy
dx
= 3x2 − 6. To find our curve f (x) we just integrate
Z
3x2 − 6 dx = x3 − 6x + C = f (x)
Now use that fact that at x = 1, y = 5 − 3(1) = 2,
so f (1) = (1)3 − 6(1) + C = 2 → C = 7
f (x) = x3 − 6x + 7
8
dy
:
dx