# Answer to Question #27113 in Integral Calculus for jan

Question #27113

find the area bounded by y=Inx,y=0 and x=e

Expert's answer

The graph of the function y=ln(x) intersects x-axis atthe point x=1.

Therefore the area between these curves is equal to

A = int_{1}^{e} (ln(x) - 0) dx =

= int_{1}^{e}ln(x) dx

Let us use integrating by parts.

Notice that

ln(x) dx = d( x ln(x) ) - x d(ln(x)) =

= d( xln(x) ) - x * 1/x

= d( xln(x) ) - 1

Hence

A = int_{1}^{e} ln(x) dx

= x ln(x)|_{1}^{e} - int_{1}^{e} 1 dx

= e*ln(e) -1*ln(1) - (e-1) =

= e - 0 - e + 1

= 1

Therefore the area between these curves is equal to

A = int_{1}^{e} (ln(x) - 0) dx =

= int_{1}^{e}ln(x) dx

Let us use integrating by parts.

Notice that

ln(x) dx = d( x ln(x) ) - x d(ln(x)) =

= d( xln(x) ) - x * 1/x

= d( xln(x) ) - 1

Hence

A = int_{1}^{e} ln(x) dx

= x ln(x)|_{1}^{e} - int_{1}^{e} 1 dx

= e*ln(e) -1*ln(1) - (e-1) =

= e - 0 - e + 1

= 1

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