5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book April 11, 2018 15:57

266 STEP 4. Review the Knowledge You Need to Score High

6. Letu= 1 +cosx;du=−sinxdxor
   −du=sinxdx.
   ∫
   sinx
   √
   1 +cosx

dx=

∫

− 1

√

u

(du)

=−

∫

1

u^1 /^2

du

=−

∫

u−^1 /^2 du

=−

u^1 /^2

1 / 2

+C

=− 2 u^1 /^2 +C

=−2(1+cosx)^1 /^2 +C

∫π

0

√sinx

1 +cosx

dx=−2(1+cosx)^1 /^2

]π

0

=− 2

[

(1+cosπ)^1 /^2

−(1+cos 0)^1 /^2

]

=− 2

[

0 − 21 /^2

]

= 2

√

2

7. Letu= 4 x;du= 4 dxor
   du
   4
   =dx.
   ∫
   g(4x)dx=

∫

g(u)

du

4

=

1

4

∫

g(u)du

=

1

4

f(u)+C

=

1

4

f(4x)+C

∫ 2

1

g(4x)dx=

1

4

f(4x)]^21

=

1

4

f(4(2))−

1

4

f(4(1))

=

1

4

f(8)−

1

4

f(4)

8.

∫ln 3

ln 2

10 exdx= 10 ex

]ln 3

ln 2

= 10

[(

eln 3

)

−

(

eln 2

)]

=10(3−2)= 10

9. Letu=t+3;du=dt.
   ∫
   1
   t+ 3
   dt=

∫

1

u

du=ln|u|+C

=ln|t+ 3 |+C

∫e 2

e

1

t+ 3

dt=ln|t+ 3 |]e

2

e

=ln (e^2 +3)−ln(e+3)

=ln

(

e^2 + 3

e+ 3

)

10. f′(x)=tan^2 x;

f′

(

π

6

)

=tan^2

(

π

6

)

=

(

1

√

3

) 2

=

1

3

11. Letu=x^2 ;du= 2 xdxor
    du
    2
    =xdx.
    ∫
    4 xex
    2
    dx= 4

∫

eu

(

du

2

)

= 2

∫

eudu= 2 eu+c= 2 ex^2 +C

∫ 1

− 1

4 xex^2 dx= 2 ex^2

] 1

− 1

= 2

[

e(1)

2

−e(−1)

2 ]

=2(e−e)= 0

Note thatf(x)= 4 xex^2 is an odd function.

Thus,

∫a

−a

f(x)dx=0.

12.

∫π

−π

(

cosx−x^2

)

dx=sinx−

x^3

3

]π

−π

=

(

sinπ−

π^3

3

)

−

(

sin(−π)−

(−π)^3

3

)

=−

π^3

3

−

(

0 −

−π^3

3

)

=−

2 π^3

3