Calculus: Area Under the Curve...Need Help!?!?

Question

Okay...I have this problem that I need help on solving...

1. a) Use rectangles to estimate the area under the curve y=x^(2) + 1, using 4 rectangles. Do so using the left endpoint of each interval.

b) Is your estimate an "overestimate" or an "underestimate" of the true area underneath the curve y= x^(2) + 1?

c) Determine the "actual area" under the curve y=x^(2) + 1 by evaluating the integral on the interval [0,1].

If anyone can...Please help me...Thanks in advance...

Answer 1

I assume the x interval is [0,1] for all of these.
Part a. Construct a table
Rectangle...xl......xr....y(xl)......y(xl)delx.....
1 0 0.25 1 0.25
2 0.25 0.50 1.0625 0.266
3 0.5 0.75 1.25 0.312
4 0.75 1.00 1.56 0.39
Sum of rectangles= 1.22
This is an underestimate because the lowest y-value in each rectangular interval is being used

The actual area is the integral; x^3/3+x+c1
c1=1, since at x=0, y=1
At 1, Int(1)=2.333 Integral= 1.333 or Int(1)-Int(0)

Answer 2

y = x² + 2x - 3 it somewhat is a parabola. the backside factor is the place the 1st by-product = 0. it somewhat is at y = -4 and x = -one million. The curve intersects the x axis the place y = 0. i.e., x² + 2x - 3 = 0. it somewhat is a quadratic and the strategies are x = -3 and x = one million (y = 0). The critical is one million/3 x^3 + x² - 3x + C Your curve intersect the x axis at -3 and +one million. it somewhat is concave upward (like a cup). while y = 0, x = -3 and + one million, so in case you combine between those 2 factors, you get in case you substitute one million, you have: one million/3 (one million)^3 + one million² - 3 X (one million) = -one million 2/3 substitute -3 one million/3 (-3)^3 + -3² - 3 X -3 = -9 Subtract perfect from backside: -32/3 the section is below the x axis, and the section is unfavorable.

Answer 3

(b-a)/(3n)[f(x0)+4f(x1)+2f(x2)+4f(x3)+f(x4)] where x0 is start; x1 is next endpoint, x2 is the next and so so on until x4 is the final endpoint
(1-0)/(3*4)*[f(0)+4f(1/4)+2f(1/2)+4f(3/4)+f(1)]
1/12*[1+4(17/16)+2(5/4)+4(25/16)+2]
1/12*[ 4/4+17/4+10/4+25/4+8/4]
1/12*[64/4]=1/12*(16)=4/3
intergrate from 0 to 1 (x^2+1) dx
x^3/3+x from 0 to 1= (x^3+3x)/3 from 0 to 1= {4/3-0}=4/3