Use the Fundamental Theorem of Calculus to evaluate the following definite integrals?

Question

I have an image of the problems, since it is not possible to write them here. Please explain a rule ( if there is one), and demonstrate how to solve the problem.

Questions are here: From top to bottom, a), b), c)

http://img262.imageshack.us/img262/8177/evaluatecalcme2.png

Answer 1

1)

∫(x^3+1)^2 dx

expand

∫(x^6 + 2x^3 + 1)dx

∫x^6 dx + 2∫x^3 dx + ∫dx

x^7/7 + 2 (x^4/4) + x

(1/7(x^7) +(1/2) x^4 + x

now apply limits 0 ,1

(1/7)[1-0] + (1/2)[1-0] + [1-0]

=>1/7 + 1/2 + 1

=> (2 + 7 + 14)/14 = 23/14

2)

∫e^(-4x) dx

-1/4 ∫e^(-4x) (-4dx)

(-1/4) e^(-4x)

now apply limits (1,3)

(-1/4) [e^(- 12) - e^(-4)]

(-1/4)[(1/e^12) - (1/e^4)]

(-1/4)[(e^4 - e^8)/e^12]

(-1/4)(-0.018)

=>0.0045

c)

∫ sin(4x) dx

(1/4)∫sin(4x) (4dx)

(1/4) (- cos(4x)

(-1/4)[cos(4pi/4) - cos(0)]

(-1/4)[ -1 - 1]

=>1/2

Answer 2

9 ? ( (a million/?x)+(a million)) dx, = ? ( x^(-a million/2)+a million) 4 9 whilst u combine, u get 2x^(a million/2) + x l 4 now use the essential Theorem, plug in 9 then 4 for the 2x^(a million/2) + x ( 2(9)^(a million/2) + 9) - ( 2(4)^(a million/2) + 4) = 15 - 8 = 7 9 ? ((a million/?x)+(a million)) dx = 7 4

Answer 3

(x^3+1)^2 = x^6 +2x^3 +1
integral = x^7/7+ x^4/2 +x
= 1/7+1/2+1 = 1 9/14

intergral = -e^(-4x)/4
= -.25e^-12 +.25e^-4 = .25(e^-4 - e^-12)

integral = -.25cos(4x)
= -.25(-1) +.25(1) = .5