MA 3972-MA-Book April 11, 2018 15:57Def 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
∫ 20g(4x)dxin terms ofG(x).Answer: Letu= 4 x;
du
4
=dx.
∫
g(u)
du
4=
1
4
G(u).Thus,∫ 20g( 4 x)dx=1
4
G( 4 x)] 20=
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 )dx2.
∫ 116(x− 2 )^1 /^2 dx3.
∫ 31t
t+ 1
dt4.
∫ 60∣∣
x− 3∣∣
dx- If
∫k0(6x−1)dx=4, findk.6.
∫π0sinx
√
1 +cosxdx- Iff′(x)=g(x) andgis a continuous
function for all real values of∫ x, express
2
1
g(4x)dxin terms off.8.
∫ln 3ln 210 exdx9.
∫e 2e1
t+ 3
dt- If f(x)=
∫x−π/ 4tan^2 (t)dt, findf′(
π
6)
.11.
∫ 1− 14 xex
2
dx12.
∫π−π(
cosx−x^2)
dxPart B Calculators are allowed.- Findkif
∫ 20(
x^3 +k)
dx= 10.- Evaluate
∫ 3. 1− 1. 22 θcosθdθto the nearest
100th.- Ify=
∫x 31√
t^2 + 1 dt, find
dy
dx.
- Use a midpoint Riemann sum with four
subdivisions of equal length to find the
approximate value of∫ 80(
x^3 + 1)
dx.- Given
∫ 2− 2g(x)dx= 8and∫ 20g(x)dx=3, find