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

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Answer 1 · 2016-07-25T19:44:09

$f (x) = \frac{1}{3} (x^{2} - 1) \sqrt{x^{2} - 1} + 3 - \sqrt{3}$ $f(x)=1/3(x^2-1)sqrt(x^2-1)+3-sqrt3$ ..

Let us subst. $x^{2} - 1 = t^{2} \Rightarrow 2 x d x = 2 t d t \Rightarrow x d x = t d t$ $x^2-1=t^2 rArr 2xdx=2tdtrArrxdx=tdt$ .

Hence, $\int x \sqrt{x^{2} - 1} d x = \int \sqrt{t^{2}} t d t = \int t^{2} d t = \frac{t^{3}}{3}$ $intxsqrt(x^2-1)dx=intsqrt(t^2)tdt=intt^2dt=t^3/3$

But, $x^{2} - 1 = t^{2} \Rightarrow t = {(x^{2} - 1)}^{\frac{1}{2}}$ $x^2-1=t^2 rArr t=(x^2-1)^(1/2)$

So, $f(x)=intxsqrt(x^2-1)dx=1/3(x^2-1)^(3/2)+C..................(1)$

To determine $C$ , we are given the cond. that $f(2)=3$

By $(1)$ , then, we have, $1/3*3^(3/2)+C=3 rArr C=3-sqrt3$

Hence, $f(x)=1/3*(x^2-1)^(3/2)+3-sqrt3$ , or,

$f(x)=1/3(x^2-1)sqrt(x^2-1)+3-sqrt3$ ..

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