Help me with this integral problem...read below~?

Question

For the definite integral (x^2+1) dx (upper limit is 4; lower limit is 0)

a. approximate this integral using right end-point and n=4 rectangles (you dont have to draw it you just have to show the numerical work)

b. evaluate the integral to find exact area

Answer 1

It's hard in Yahoo answers to do (a). I can tell you how to do it. Graph the function y = x^2+1. Select three points between 0 and 4, and draw upward lines from these up to the function. Then draw horizontal lines from this point to the line drawn from the next point to make a rectangle. You should get 4 rectangles.

It's hard to do it numerically, too. You could select 0 < x1 < x2 < x3 < 4 to be your three points. I will describe a rectangle by giving two opposite corners:

(0,0) - (x1,f(x1))
(x1,0)-(x2,f(x2))
(x2,0)-(x3,f(x3))
(x3,0)-(4,17)

The area would be (x1-0)*(f(x1)) + (x2-x1)*f(x2) + (x3-x2)*f(x3) + (4-x3)*f(4)

b. The integral of x^2+1 is (1/3)x^3+x+C for C an arbitrary constant. From 0 to 4, it is

(1/3)x^3+x|(0,4) = 1/3(4^3)+4 - (1/3)(0^3)-0 = 64/3+4 = 76/3.

Answer 2

(a) n=4 ==> each rectangle has width 1 ==> approx value = 1*f(1) + 1*f(2) + 1*f(3) + 1*f(4) = 2 + 5 + 10 + 17 = 34.

(b) exact = (x^3 / 3 + x) evaluated from 0 to 4 = 64/3 + 4 = 25 1/3.

Answer 3

4
∫(x^2+1)dx
0

x.. y min mid max
0.. 1
1.. 2.. 1. 1.5.... 2
2.. 5.. 2. 3.5.... 5
3 10.. 5. 7.5.. 10
4 17 10 13.5.. 17
. . . 18. 26.... 34

4
∫(x^2+1)dx = (1/3)x^3 + x from 0 to 4 = 64/3 + 4 = 76/3
0
= 25.33333.....