MA 3972-MA-Book May 8, 2018 13:52
Integration 231Example 3
If
dy
dx
= 3 x^2 +2, and the point (0,−1) lies on the graph ofy, findy.
Since
dy
dx
= 3 x^2 +2, thenyis an antiderivative of
dy
dx. Thus,
y=
∫ (
3 x^2 + 2)
dx=x^3 + 2 x+C. The point (0,−1) is on the graph ofy.Thus,y =x^3 + 2 x+C becomes− 1 = 03 +2(0)+C orC=−1. Therefore,y =x^3 +
2 x−1.
Example 4
Evaluate
∫ (
1 −1
√ (^3) x 4
)
dx.
Rewrite as
∫ (
1 −
1
x^4 /^3)
dx=∫ (
1 −x−^4 /^3)
dx=x−
x−^1 /^3
− 1 / 3
+C=x+3
√ (^3) x+C.
Example 5
Evaluate
∫
3 x^2 +x− 1
x^2
dx.
Rewrite as
∫ (
3 +
1
x−
1
x^2)
dx=∫ (
3 +1
x
−x−^2)
dx= 3 x+ln|x|−
x−^1
− 1
+C= 3 x+ln|x|+1
x+C.
Example 6
Evaluate
∫ √
x(
x^2 − 3)
dx.Rewrite as See Example 5
∫
x^1 /^2(
x^2 − 3)
dx=∫ (
x^5 /^2 − 3 x^1 /^2)
dx=
x^7 /^2
7 / 2−
3 x^3 /^2
3 / 2+C=
2
7
x^7 /^2 − 2√
x^3 +C.Example 7
Evaluate
∫ (
x^3 −4 sinx)
dx.∫ (
x^3 −4 sinx
)
dx=
x^4
4+4 cosx+C.Example 8
Evaluate
∫
(4 cosx−cotx)dx.∫
(4 cosx−cotx)dx=4 sinx−ln|sinx|+C.