228 STEP 4. Review the Knowledge You Need to Score High
- Rewrite:
∫ (
e^4 x
ex−
1
ex)
dx=
∫ (
e^3 x−e−x)
dx=∫
e^3 xdx−∫
e−xdx.
Letu= 3 x;du= 3 dx;
∫
e^3 xdx=∫
eu(
du
3)
=1
3
eu+C 1=1
3
e^3 x+C.
Letv=−x;dv=−dx;
∫
e−xdx=∫
ev(−dv)=ev+C 2
=−e−x+C 2
Thus,∫
e^3 xdx−∫
e−xdx=1
3
e^3 x+e−x+C.Note:C 1 and C 2 are arbitrary constants,
and thusC 1 +C 2 =C.15.∫Rewrite:
(9−x^2 )x^1 /^2 dx=∫ (
9 x^1 /^2 −x^5 /^2)
dx=9 x^3 /^2
3 / 2−
x^7 /^2
7 / 2+C
= 6 x^3 /^2 −
2 x^7 /^2
7+C.
- Letu= 1 +x^3 /^2 ;du=
3
2
x^1 /^2 dxor
2
3
du=x^1 /^2 dx=√
xdx.Rewrite:∫
u^4(
2
3
du)
=2
3
∫
u^4 du=
2
3
(
u^5
5)
+C=2
(
1 +x^3 /^2)155
+C.- Since
dy
dx
=ex+2, theny=
∫
(ex+ 2 )dx=ex+ 2 x+C.
The point (0, 6) is on the graph ofy.
Thus, 6=e^0 +2(0)+C⇒ 6 = 1 +Cor
C=5. Therefore,y=ex+ 2 x+5.- Letu=ex;du=exdx.
Rewrite:− 3
∫
sin(u)du=−3(−cosu)+C=3 cos(ex)+C.- Letu=ex+e−x;du=(ex−e−x)dx.
Rewrite:
∫
1
u
du=ln|u|+C=ln|ex+e−x|+C
or=ln∣∣
∣∣ex+^1
ex∣∣
∣∣+C=ln∣∣
∣∣e2 x+ 1
ex∣∣
∣∣+C=ln|e^2 x+ 1 |−ln|ex|+C
=ln|e^2 x+ 1 |−x+C.- Sincef(x) is the antiderivative of
1
x,
f(x)=∫
1
x
d=ln|x|+C.Givenf(1)=5; thus, ln (1)+C= 5
⇒ 0 +C=5orC=5.
Thus,f(x)=ln|x|+5 and
f(e)=ln (e)+ 5 = 1 + 5 =6.- Integrate
∫
x^2√
1 −xdxby parts. Letu=x^2 ,du= 2 xdx,
dv=√
1 −xdx, andv=−2
3
(1−x)^3 /^2. Then
∫
x^2√
1 −xdx=−2
3
x^2 ( 1 −x)^3 /^2+4
3
∫
x(1−x)^3 /^2 dx.Use parts again with
u=x,du=dx,dv=(1−x)^3 /^2 dx,
andv=−2
5
( 1 −x)^5 /^2 so that
∫
x^2√
1 −xdx=−2
3
x^2 (1−x)^3 /^2+4
3
[
−2
5
(1−x)^5 /^2 −∫
−2
5
( 1 −x)^5 /^2 dx]
.Integrate for−2
3
x^2 ( 1 −x)^3 /^2 −8
15
x