Finding the Area Bounded by Two Graphs In Exercises 15-30, sketch the region bounded by the graphs of the functions and find the area of the region. See Examples 1, 2, 3, and 4.

y = x 2 − 2 x + 1 , y = x 2 − 10 x + 25 , y = 0

To graph: The region bounded by the graphs of y=x2−2x+1, y=x2−10x+25 and y=0, and also compute the area of the region.

Given Information:

The region bounded by the graphs of y=x2−2x+1, y=x2−10x+25 and y=0.

Graph:

Consider the following equations that give the required region,

y=x2−2x+1, y=x2−10x+25 and y=0

It is a quadratic function. The term x2 is positive, so its graph will be a parabola with opening upwards. Compute its x-intercepts as follows:

x2−2x+1=0(x−1)2=0x−1=0x=1

The x-intercept is x=1.

Compute its y-intercepts as follows:

y=02−2×0+1y=1

The y-intercept is y=1.

Consider the second function y=x2−10x+25

It is a quadratic function. The term x2 is positive, so its graph will be a parabola with opening upwards. Compute its x-intercepts as follows:

x2−10x+25=0(x−5)2=0x−5=0x=5

The x-intercept is x=5.

Compute its y-intercepts as follows:

y=02−10×0+25y=25

The y-intercept is y=25.

Compute the intersection point of the graphs of y=x2−2x+1 and y=x2−10x+25 as follows:

x2−2x+1=x2−10x+2510x−2x=25−18x=24x=3

Substitute x=3 in the equation y=x2−2x+1 and compute the intersection point,

y=32−2×3+1=9−6+1=4

So two graphs intersect at point (3,4).

Compute the x-intercept point of the graph of y=x2−2x+1 as follows:

x2−2x+1=0(x−1)2=0x−1=0x=1

So, x-intercept of the graph of y=x2−2x+1 is (1,0).

Compute the x-intercept point of the graph of y=x2−10x+25 as follows:

x2−10x+25=0(x−5)2=0x−5=0x=5

So, x-intercept of the graph of y=x2−10x+25 is (5,0).

Intersection point is calculated by,

x2−10x+25=x2−2x+1x=3

Sketch the graph of the region as follows:

Formula used:

Area of a region bounded by two graphs is calculated using the following formula,

If f and g are continuous on [a,b] and g(x)≤f(x) for all x in [a,b], then the area of the region bounded by the graphs of f, g, x=a and x=b is given by

A=∫ab[f(x)−g(x)]dx

Power for integrals is,

∫xndx=xn+1n+1+C

Calculate:

From the above, the required area consists of two regions, left region and right region