FIND INTEGRAL 7^3 SQUARE ROOT X

Question

How to find integral of ∫7³√x

is there a way you can explain

 

Answer 1

 

I think he means : ∫7^{3√X =} 7 ∫ X  ^{1/3 =}      7          X ^4/3    + C 

                                                           1 + 1/3 

                                                  

                                                  = 21  X ^4/3  + C

                                                      4

Answer 2

∫7³√x dx

 

We can take the constant (7³) outside of the integral sign.  Also change √x to x^1/2.

 

= 7³∫x^1/2 dx

 

= 7³*(2/3)*x^3/2 + C,  where C is a constant of integration

 

The anti-derivative of x^1/2 is (2/3)x^3/2.  To confirm, work the power rule for a derivative backwards.

 

Power rule: d(axⁿ)/dx = anx^n-1  where a is a constant

 

Normally, you're given axⁿ and are asked to find its derivative.  In this case, we're given the derivative, anx^n-1 (=x^1/2), and we want to find the original function, anxⁿ, or anti-derivative.

 

So we want anx^n-1 = x^1/2

 

n-1 = 1/2, therefore n = 3/2

 

an = 1 = a*(3/2), therefore a=2/3

 

axⁿ = (2/3)x^3/2  (this is the anti-derivative of x^1/2)