How Can You Effectively Evaluate a Challenging Definite Integral?

[anemone]

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

 

[MarkFL]

My solution:

Spoiler

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:

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

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

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

 

[anemone]

My solution:

Spoiler

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)

 

[Mathwebster]

A little more direct

Spoiler

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 $I = 1$

 

[CellCoree]

There are a few different approaches one could take to evaluate this definite integral. One possible method is to use the substitution u = ln(9-x), which would result in the integral becoming \int_{ln7}^{ln5} \frac{\sqrt{u}\,du}{\sqrt{u}+\sqrt{ln(12-e^u)}}. From here, one could potentially use integration by parts or another technique to further simplify the integral.

Another approach could be to use a numerical method, such as the trapezoidal rule or Simpson's rule, to approximate the value of the integral. This would involve breaking the interval [2,4] into smaller subintervals and using a formula to calculate the area under the curve on each subinterval.

Ultimately, the method chosen would depend on the specific goals and context of the problem. It may also be helpful to graph the integrand and visually estimate the area under the curve, or to use a computer program or calculator to numerically evaluate the integral.