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

Tushar Chandra | Certified Educator

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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)

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## Related Questions

giorgiana1976 | Student

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.