   Chapter 7.1, Problem 57E

Chapter
Section
Textbook Problem

# Find the area of the region bounded by the given curves. y = x 2 ln x ,     y = 4 ln x

To determine

To evaluate: area of the region bounded by two given curves

Explanation

Consider a function y=f(x) between the points x=a and x=b. The area under this graph between the two points is given by the following integral:

A=abydx

Then, the area between two curves will be given by the difference between the area under their respective graphs.

Formula used:

Area between two curves f and g is given by:

A=ab|(f(x)g(x))|dx …… (1)

Given:

The two curves, y=x2lnx,y=4lnx.

Calculation:

Plot the two curves together to get a visual sense of the area to be calculated:

In the graph above, red curve is y=x2lnxand the blue curve is y=4lnx.

The graph shows that the two curves intersect at the points (1,0) and (2,2.773).

Find the point of intersection numerically by equating the two curves:

x2lnx=4lnxx2=4x=2

Here, the negative root was ignored as the domain of the function is of positive real numbers. Since, ln1 is zero, the equation above is satisfied for x=1. Hence, the graph intersects at x=1 and x=2, same as we found from the graphical method.

So, the limits of the integral a and b to find area using formula (1) will be 1 and 2 respectively.

Substitute the curves and the limit of integration into the formula to get;

A=12|(x2lnx4lnx)|dx

In the region 1x2, the graph of 4lnx is greater than x2lnx, so

|(x2lnx4lnx)|=(4lnxx2lnx)      1x2

So, the area will be:

A=12|(x2lnx4lnx)|dx=12(4lnxx2lnx)dx=124lnxdx12x2lnxdx

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