What is #f(x) = int xsqrt(3x) dx# if #f(3) = 0 #?

Question

What is #f(x) = int xsqrt(3x)   dx# if #f(3) = 0  #?

Cameron Alexander · Accepted Answer

#=>f(x) = (2sqrt(3))/5x^(5/2) -54/5# #f(x) = int xsqrt(3x) dx# #=>f(x) = sqrt(3)int xsqrt(x) dx# #=>f(x) = sqrt(3)int x^(3/2) dx# #=>f(x) = sqrt(3)x^(3/2+1)/(3/2+1) +c# [where c = Integration constant] #=>f(x) = (2sqrt(3))/5x^(5/2) +c.....(1)# Again given condition is #f(3)=0# So #=>f(3) = (2sqrt(3))/5xx3^(5/2) +c# #=>0= 2/5xx3^(5/2+1/2) +c# #=>0= 2/5xx3^3 +c# #=>0= 54/5 +c# #=>c= -54/5 # Hence we have , substituting the value of c in eq(1) #=>f(x) = (2sqrt(3))/5x^(5/2) -54/5#

Ezekiel Alexander · Answer

undefined To find the function $ f(x) = \int x\sqrt{3x} \, dx $ when $ f(3) = 0 $, we can first find the antiderivative of $ x\sqrt{3x} $ and then evaluate it at $ x = 3 $.

First, let's find the antiderivative:

$$ \int x\sqrt{3x} \, dx $$

Let $ u = 3x $, then $ du = 3dx $ or $ dx = \frac{du}{3} $.

$$ \int x\sqrt{3x} \, dx = \int \sqrt{u} \cdot \frac{u}{3} \, du = \frac{1}{3} \int u^{\frac{3}{2}} \, du $$

Using the power rule for integration:

$$ \frac{1}{3} \cdot \frac{2}{5}u^{\frac{5}{2}} + C = \frac{2}{15}u^{\frac{5}{2}} + C $$

Substitute back $ u = 3x $:

$$ \frac{2}{15}(3x)^{\frac{5}{2}} + C = \frac{2}{15} \cdot 3^{\frac{5}{2}}x^{\frac{5}{2}} + C $$

Now, we know that $ f(3) = 0 $, so we can plug in $ x = 3 $ and solve for $ C $:

$$ \frac{2}{15} \cdot 3^{\frac{5}{2}} \cdot 3^{\frac{5}{2}} + C = 0 $$
$$ \frac{2}{15} \cdot 3^5 + C = 0 $$
$$ \frac{2}{15} \cdot 243 + C = 0 $$
$$ \frac{486}{15} + C = 0 $$
$$ C = -\frac{486}{15} $$

So, the function is:

$$ f(x) = \frac{2}{15} \cdot 3^{\frac{5}{2}}x^{\frac{5}{2}} - \frac{486}{15} $$