What is $f (x) = \int x \sqrt{x^{2} - 1} d x$ $f(x) = int xsqrt(x^2-1) dx$ if $f (3) = 0$ $f(3) = 0$ ?

Question

1305 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-02T12:36:06

$f (x) = \frac{1}{3} {(x^{2} - 1)}^{\frac{3}{2}} - \frac{16 \sqrt{2}}{3}$ $f(x) = 1/3(x^2-1)^(3/2) - (16 sqrt 2)/(3)$

$\int x \sqrt{x^{2} - 1} d x$ $int xsqrt(x^2-1) dx$

$= int \ d/dx ( 2/3* 1/2(x^2-1)^(3/2) ) dx$

$= int \ d/dx ( 1/3(x^2-1)^(3/2) ) dx$

$= 1/3(x^2-1)^(3/2) + C$

using the IV:

$0 = 1/3(3^2-1)^(3/2) + C implies C = - (16 sqrt 2)/(3)$

So $f(x) = 1/3(x^2-1)^(3/2) - (16 sqrt 2)/(3)$

What is f(x) = int xsqrt(x^2-1) dxf(x)=∫x√x2−1dxf(x) = int xsqrt(x^2-1) dx if f(3) = 0 f(3)=0f(3) = 0 ?