Help -find the total area of the region between the curve and the x-axis

y= x^2 + 7;
4
∫
2

y=-x^2+9;
5
∫
0

y=(x+1)^3;

3
∫
0


thanks !

I'm not going to do your homework for you (that's what Wolfram|Alpha is for). If there's a problem you're having with integration techniques or something, then I can help with that. Otherwise, get someone else to do your work.

∫ {2, 4} (x^2 + 7) dx = x^3/3 + 7x

Evaluating: (4^3/3 + 7 • 4) - (2^3/3 + 7 • 2) = 64/3 + 28 - 8/3 - 14 = 56/3 + 14 = 98/3

∫ {0, 5} (-x^2 + 9) dx = -x^3/3 + 9x

Evaluating: (-5^3/3 + 9 • 5) - (0^3/3 + 9 • 0) = -125/3 + 45 = 10/3

∫ {0, 3} (x + 1)^3 dx = (x + 1)^4/4

Evaluating: (4^4/4) - (1^4/4) = 64 - 1/4 = 255/4

These are just regular integrals.

so like for the first one, the integral of x^2+7 with respect to x is (x^3)/3+7x.
now just put in the bounds 2 and 4,
(4^3)/3+7*4-(2^3)/3-7*2=98/3

If you don't know how to do integrals you should consult your teacher. If you just don't want to do the math, use Wolfram Alpha
http://www.wolframalpha.com/input/?i=int…

1.
∫(x^2 + 7)dx [2,4]
=(1/3)x^3 + 7x +C [2,4]
=[(1/3)(4^3)+7(4)]-[(1/3)(2^3) + 7(2)]
=98/3

2.
∫(-x^2+9)dx [0,5]
=-∫x^2dx +∫9dx [0,5]
=-(1/3)(x^3)+9x [0,5]
=[-(1/3)(5^3)+9(5)]
=10/3

3.
∫(x+1)^3dx [0,3]
let u=x+1, du=dx
∫u^3du
=(1/4)u^4
=(1/4)(x+1)^4 [0,3]
=[(1/4)(3+1)^4] - [(1/4)(0+1)^4]
=255/4

Hi, sorry but it seems that that question would most likely come rom your homework, and you just copied that from your homework so you just want someone else to answer it for you. Maybe look it up in your math book, nd look at it step by step.
Hope this helps! :)