another integral

**Random Variable** · August 23rd 2009, 11:28 AM

$\int_{3}^{4} \frac{x^{\frac{1}{7}}}{(7-x)^{\frac{1}{7}} + x^{\frac{1}{7}}} \ dx$

It's extremely simple if you spot the trick (which I didn't).

**Chris L T521** · August 23rd 2009, 11:50 AM

Originally Posted by Random Variable

$\int_{3}^{4} \frac{x^{\frac{1}{7}}}{(7-x)^{\frac{1}{7}} + x^{\frac{1}{7}}} \ dx$

It's extremely simple if you spot the trick (which I didn't).

Let $I_1=\int_3^4\frac{\left(7-x\right)^{\frac{1}{7}}+x^{\frac{1}{7}}}{\left(7-x\right)^{\frac{1}{7}}+x^{\frac{1}{7}}}\,dx=4-3=1$ .

Now,

$I_2=\int_3^4\frac{x^{\frac{1}{7}}}{\left(7-x\right)^{\frac{1}{7}}+x^{\frac{1}{7}}}\,dx$

Let $u=7-x\implies -\,du=\,dx$ .

So $I_2=\int_3^4\frac{x^{\frac{1}{7}}}{\left(7-x\right)^{\frac{1}{7}}+x^{\frac{1}{7}}}\,dx\xright arrow{u=7-x}{}\int_3^4\frac{\left(7-u\right)^{\frac{1}{7}}}{u^{\frac{1}{7}}+\left(7-u\right)^{\frac{1}{7}}}\,du$ $=\int_3^4\frac{u^{\frac{1}{7}}+\left(7-u\right)^{\frac{1}{7}}}{u^{\frac{1}{7}}+\left(7-u\right)^{\frac{1}{7}}}\,du-\int_3^4\frac{u^{\frac{1}{7}}}{u^{\frac{1}{7}}+\le ft(7-u\right)^{\frac{1}{7}}}\,du$

Therefore,

$2I_2=I_1=1\implies I_2=\frac{1}{2}$

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