What is #f(x) = int xsqrt(x-2) dx# if #f(2) = 3 #?

Question

What is #f(x) = int xsqrt(x-2)   dx# if #f(2) = 3   #?

Isabella Ackerman · Accepted Answer

#f(x)=(2x(x-2)^(3/2))/3-(4(x-2)^(5/2))/15+3# #f(x)=int\ xsqrt(x-2)\ dx# According to integration by parts, #int\ u\ dv=uv-int\ v\ du#. Set #u=x# and #dv=sqrt(x-2)\ dx=(x-2)^(1/2)\ dx#. Then, #du=dx# and #v=(2(x-2)^(3/2))/3#. Thus, #f(x)# #=int\ xsqrt(x-2)\ dx# #=(2x(x-2)^(3/2))/3-int\ (2(x-2)^(3/2))/3\ dx# #=(2x(x-2)^(3/2))/3-(4(x-2)^(5/2))/15+C# Now, since #f(2)=3#, #(2*2(2-2)^(3/2))/3-(4(2-2)^(5/2))/15+C=3# #C=3# Thus, the final answer is #f(x)=(2x(x-2)^(3/2))/3-(4(x-2)^(5/2))/15+3#

Axel Alvarez · Answer

undefined To find $ f(x) = \int x\sqrt{x - 2} \, dx $ when $ f(2) = 3 $, we need to solve the integral using integration by parts.

Let $ u = x $ and $ dv = \sqrt{x - 2} \, dx $.

Then, $ du = dx $ and $ v = \frac{2}{3}(x - 2)^\frac{3}{2} $.

Using the integration by parts formula:
$$ \int u \, dv = uv - \int v \, du $$

$$ \int x\sqrt{x - 2} \, dx = \frac{2}{3}x(x - 2)^\frac{3}{2} - \int \frac{2}{3}(x - 2)^\frac{3}{2} \, dx $$

$$ = \frac{2}{3}x(x - 2)^\frac{3}{2} - \frac{2}{5}(x - 2)^\frac{5}{2} + C $$

Given that $ f(2) = 3 $, we can substitute $ x = 2 $ into the equation and solve for $ C $:
$$ 3 = \frac{2}{3}(2)(2 - 2)^\frac{3}{2} - \frac{2}{5}(2 - 2)^\frac{5}{2} + C $$
$$ 3 = 0 - 0 + C $$
$$ C = 3 $$

So, $ f(x) = \frac{2}{3}x(x - 2)^\frac{3}{2} - \frac{2}{5}(x - 2)^\frac{5}{2} + 3 $.