Find the area of the region enclosed by y=ln(x) ,the x-axis,the y-axis and y=1 ? (a) dx select (b) dy select

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2 Answers
Parabola
Apr 7, 2018

#A=e-1#

Explanation:

I have the graphs of the functions and lines here:

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We want the area of the green region.

We can think of the situation like this:

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Now, let's find the intercept and the intersection.

The intercept:

#ln(x)=0#

#=>x=e^0#

#=>x=1#

The intersection:

#1=ln(x)#

#e^1=x#

#e=x#

We can now form a rectangle like the following:

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The area of the green region is the area of the rectangle minus the area under the curve of #ln(x)# from #1# to #e#

#A=e*1-int_1^eln(x)dx#

The fundamental theorem of calculus:

#int_a^bf(x)dx=F(b)-F(a)# if #F'(x)=f(x)#

Remember that #intln(x)dx=x(ln(x)-1)+C#

#=>A=e-int_1^eln(x)dx#

#=>A=e-(e(ln(e)-1)-1(ln(1)-1))#

#=>A=e-(e(1-1)-1(0-1))#

#=>A=e-(e(0)-1(-1)#

#=>A=e-(0+1)#

#=>A=e-1#

#=>A~~ 1.71828182846#

In case you are wondering how I solved #intln(x)dx#...

Integration by parts:

#intudv=uv-intvdu#

#u=ln(x)#

#dv=1#

#=>du=d/dx(ln(x))#

#=>du=1/x#

#=>v=int1dx#

#=>v=x#

#=>xln(x)-intx*1/xdx#

#=>xln(x)-int1dx#

#=>xln(x)-x#

#=>x(ln(x)-1)#

Sean
Apr 7, 2018

#color(red)(A=e-1)#

Explanation:

.

Let's take a look at the graph of this area:

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The area described in the problem is shaded in yellow. If we use integration along the #x#-axis, we will have to set up our integral to calculate the area between the green line, #(y=1)# and the red curve #(y=lnx)#.

This will give us the combination of the yellow and pink areas.

Then we will set up another integral to calculate the area between the #x#-axis, #(y=0)# and the red curve #(y=lnx)#; and subtract it from the combined area to get the yellow area. In each case, the argument of the integral will be the upper function minus the lower function. We will integrate between #x=0# and #x=# the intersection of the green and red curves.

To calculate the #x#-coordinate of the point of intersection, we set the two functions equal to each other and solve for #x#:

#lnx=1# or when written in exponential form #e^1=x, :. x=e#

#A_("Combined")=int_0^e(1-lnx)dx=x-xlnx+x=2x-xlnx#

#A_("Combined")=(x(2-lnx))_0^e=e#

#A_("Pink")=int_0^1(0-lnx)dx=(-xlnx+x)_0^1=1#

#A_("Yellow")=A_("Combined")-A_("Pink")=e-1#

If we use integration along the #y#-axis, we have to convert our functions to #x# as a function of #y# first:

#y=lnx# when written in exponential form becomes #x=e^y# by the definition of logarithms.

Now, we can use one integral to calculate the area between the #y#-axis #(x=0)# and the red curve #(x=e^y)# between #0# and #1#:

#A_("Yellow")=int_0^1(e^y-0)dy=(e^y)_0^1=e-1#

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