What is $f (x) = \int x \sqrt{5 - 2} d x$ $f(x) = int xsqrt(5-2) dx$ if $f (2) = 3$ $f(2) = 3$ ?

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Answer 1 · 2018-04-29T06:34:00

The function $f (x)$ $f(x)$ is $\frac{x^{2} \sqrt{3} + 6 - 4 \sqrt{3}}{2}$ $(x^2sqrt3+6-4sqrt3)/2$ .

First, compute the integral:

$f (x) = \int x \sqrt{5 - 2}$ $f(x)=intxsqrt(5-2)$ $d x$ $dx$

$f (x) = \int x \sqrt{3}$ $color(white)(f(x))=intxsqrt3$ $d x$ $dx$

$f (x) = \sqrt{3} \int x$ $color(white)(f(x))=sqrt3intx$ $d x$ $dx$

$f (x) = \sqrt{3} \cdot \frac{x^{2}}{2} + C$ $color(white)(f(x))=sqrt3*x^2/2+C$

Now, set $f (2)$ $f(2)$ (which is $3$ $3$ ) and the integral evaluated at $2$ $2$ equal to each other, then solve for $C$ $C$ :

$3 = \sqrt{3} \cdot \frac{2^{2}}{2} + C$ $3=sqrt3*2^2/2+C$

$3 = 2 \sqrt{3} + C$ $3=2sqrt3+C$

$3 - 2 \sqrt{3} = C$ $3-2sqrt3=C$

That's the $C$ $C$ value, so that means that the function is:

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

If you would like, you can rewrite it as:

$f (x) = \frac{x^{2} \sqrt{3} + 6 - 4 \sqrt{3}}{2}$ $f(x)=(x^2sqrt3+6-4sqrt3)/2$

Hope this helped!

What is f(x) = int xsqrt(5-2) dxf(x)=∫x√5−2dxf(x) = int xsqrt(5-2) dx if f(2) = 3 f(2)=3f(2) = 3 ?