Welcome to our community

Be a part of something great, join today!

Definite Integral Challenge

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,680
Evaluate \(\displaystyle \int_2^4 \frac{\sqrt{\ln (9-x)}\,dx}{\sqrt{\ln (9-x)}+\sqrt{\ln (x+3)}}\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
My solution:

We are given to evaluate:

\(\displaystyle I=\int_2^4 \frac{\sqrt{\ln (9-x)}\,dx}{\sqrt{\ln (9-x)}+\sqrt{\ln (x+3)}}\)

Let:

\(\displaystyle u=x-3\,\therefore\,du=dx\)

and we have:

\(\displaystyle I=\int_{-1}^{1}\frac{\sqrt{\ln(6-u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}\,du\)

Using the property of definite integrals:

\(\displaystyle \int_{-a}^a f(x)\,dx=\int_0^a \left[f(x)+f(-x) \right]\,dx\)

we may write:

\(\displaystyle I=\int_{0}^{1}\frac{\sqrt{\ln(6-u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}+\frac{\sqrt{\ln(6+u)}}{\sqrt{\ln(6+u)}+\sqrt{\ln(6-u)}}\,du\)

\(\displaystyle I=\int_{0}^{1}\frac{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}\,du\)

\(\displaystyle I=\int_{0}^{1}\,du=_0^1=1-0=1\)
 
  • Thread starter
  • Admin
  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,680
My solution:

We are given to evaluate:

\(\displaystyle I=\int_2^4 \frac{\sqrt{\ln (9-x)}\,dx}{\sqrt{\ln (9-x)}+\sqrt{\ln (x+3)}}\)

Let:

\(\displaystyle u=x-3\,\therefore\,du=dx\)

and we have:

\(\displaystyle I=\int_{-1}^{1}\frac{\sqrt{\ln(6-u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}\,du\)

Using the property of definite integrals:

\(\displaystyle \int_{-a}^a f(x)\,dx=\int_0^a \left[f(x)+f(-x) \right]\,dx\)

we may write:

\(\displaystyle I=\int_{0}^{1}\frac{\sqrt{\ln(6-u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}+\frac{\sqrt{\ln(6+u)}}{\sqrt{\ln(6+u)}+\sqrt{\ln(6-u)}}\,du\)

\(\displaystyle I=\int_{0}^{1}\frac{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}\,du\)

\(\displaystyle I=\int_{0}^{1}\,du=_0^1=1-0=1\)


That's a smart approach, MarkFL! Well done!(Clapping)
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
A little more direct
Let $x = 6-u$ so the integral become

$I = \displaystyle - \int_4^2 \dfrac{\sqrt{\ln (3+u)}}{ \sqrt{\ln (3+u)} +\sqrt{\ln (9-u)}}\;du$

Replace $u$ with $x$ and add to the original and like Mark said $2I = \int_2^4 1dx$ gives $I = 1$