MA 3972-MA-Book April 11, 2018 15:57
Def inite Integrals 263
- IfG(x) is an antiderivative of (ex+1) andG(0)=0, findG(1).
Answer: G(x)=ex+x+C
G(0)=e^0 + 0 +C= 0 ⇒C=−1.
G(1)=e^1 + 1 − 1 =e. - IfG′(x)=g(x), express
∫ 2
0
g(4x)dxin terms ofG(x).
Answer: Letu= 4 x;
du
4
=dx.
∫
g(u)
du
4
=
1
4
G(u).Thus,
∫ 2
0
g( 4 x)dx=
1
4
G( 4 x)
] 2
0
=
1
4
[G(8)−G(0)].
12.5 Practice Problems
Part A The use of a calculator is not
allowed.
Evaluate the following definite integrals.
1.
∫ 0
− 1
(1+x−x^3 )dx
2.
∫ 11
6
(x− 2 )^1 /^2 dx
3.
∫ 3
1
t
t+ 1
dt
4.
∫ 6
0
∣∣
x− 3
∣∣
dx
- If
∫k
0
(6x−1)dx=4, findk.
6.
∫π
0
sinx
√
1 +cosx
dx
- Iff′(x)=g(x) andgis a continuous
function for all real values of∫ x, express
2
1
g(4x)dxin terms off.
8.
∫ln 3
ln 2
10 exdx
9.
∫e 2
e
1
t+ 3
dt
- If f(x)=
∫x
−π/ 4
tan^2 (t)dt, findf′
(
π
6
)
.
11.
∫ 1
− 1
4 xex
2
dx
12.
∫π
−π
(
cosx−x^2
)
dx
Part B Calculators are allowed.
- Findkif
∫ 2
0
(
x^3 +k
)
dx= 10.
- Evaluate
∫ 3. 1
− 1. 2
2 θcosθdθto the nearest
100th.
- Ify=
∫x 3
1
√
t^2 + 1 dt, find
dy
dx
.
- Use a midpoint Riemann sum with four
subdivisions of equal length to find the
approximate value of
∫ 8
0
(
x^3 + 1
)
dx.
- Given
∫ 2
− 2
g(x)dx= 8
and
∫ 2
0
g(x)dx=3, find