1
$\begingroup$

$$\int_0^3 \frac{x^4}{x^4+(x-3)^4} \,dx$$

Then the question ask me to change it into

$$\int_0^3 \frac{(x-3)^4}{x^4+(x-3)^4} \,dx$$

Then how to evaluate it

if let $$u=x-3$$ $$du=dx$$ $$x^4 =(u+3)^4$$ $$\int_3^0 \frac{u^4}{(u+3)^4+u^4} \,du$$ it is still the same so what should i do?

  • 1
    $$\int_a^b dx = -\int_b^a dx$$ Have you seen this? – 2017-02-01
  • 0
    See http://math.stackexchange.com/questions/2121087/evaluation-of-int-pi-2-0-sqrt-sin-2x-cdot-sinx-cdot-dx/2121124#2121124 – 2017-02-01
  • 1
    You need to use $u= 3-x$; not $u=x-3$. Then add the result of that substitution to the original. – 2017-02-01
  • 2
    I do not understand why down-vote. Its not difficult, but efforts were shown. – 2017-02-01

2 Answers 2

2

Let,

$$I=\int_0^3 \frac{x^4}{x^4+(x-3)^4} \,dx$$

Let $x=3-u$ and change dummy variable back to $x$.

$$I=\int_{0}^{3} \frac{(3-x)^4}{x^4+(x-3)^4} dx$$

$$=\int_{0}^{3} \frac{(x-3)^4}{x^4+(x-3)^4} dx$$

Add this to the first form of $I$.

$$2I=\int_{0}^{3} 1 dx$$

  • 0
    oh thanks seems like that i haven't woken up – 2017-02-01
1

As suggested

as $\displaystyle I=\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$

$$2I=\int_a^b[f(x)+f(a+b-x)]\ dx$$

Here $f(3+0-x)=\dfrac{(3-x)^4}{(3-x)^4+x^4}$

$$2I=\int_0^3dx$$