Question

Definite integral of x^2/sqrt(x^3+1), between 2 and 3, gives=?

Accepted Answer

We have to find the value of the definite integral of x^2/sqrt (x^3 + 1) between the limits x = 2 and x = 3.
First we determine the indefinite integral and then substitute the values x = 3 and x = 2.
Int [ x^2 / sqrt (x^3 + 1) dx ]
let y = x^3 + 1
dy/3 = x^2*dx
=> Int[(1/3)*y^(-0.5) dy]
=> (2/3)*sqrt y + C
substitute y = x^3 + 1
=> (2/3)*sqrt (x^3 + 1) + C
Between the limits x = 2 and x = 3
(2/3)*sqrt (3^3 + 1) + C - (2/3)*sqrt (2^3 + 1) - C
=> (2/3)(sqrt 28 - sqrt 9)
=> (2/3)(2*sqrt 7 - 3)
The required value of the definite integral is (2/3)(2*sqrt 7 - 3)

Answer

We'll apply Leibniz-Newton formula to evaluate the definite integral:
Int f(x)dx = F(b) - F(a), where a and b are the limits of integration.
First, we'll determine the indefinite integral. We'll change the variable as method of solving the integral.
Int x^2dx/sqrt(x^3 + 1)
We notice that if we'll put x^3 + 1 = t and we'll differentiate, we'll get the numerator.
3x^2dx = dt
x^2dx = dt/3
We'll re-write the integral:
Int x^2dx/sqrt(x^3 + 1) = Int (dt/3)/sqrt t
Int (dt/3)/sqrt t = (1/3)*Int dt/sqrt t
(1/3)*Int dt/sqrt t = (1/3)*[t^(-1/2 + 1)/(-1/2 + 1)] + C
(1/3)*Int dt/sqrt t = (2/3)*sqrt t + C
Int f(x)dx = (2/3)*sqrt (x^3 + 1) + C
We'll evaluate the definite integral, having as limits of integration x = 2 and x = 3:
Int f(x)dx = (2/3)*sqrt (3^3 + 1) - (2/3)*sqrt (2^3 + 1)
Int f(x)dx = (2/3)*(sqrt28 - 3)
Int f(x)dx = (2*sqrt28)/3 - 2
The definite integral of the function f(x) = x^2/sqrt(x^3 + 1), is Int f(x)dx = (2*sqrt28)/3 - 2.

Calculus

Definite integral of x^2/sqrt(x^3+1), between 2 and 3, gives=?

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