Calculate the definite integral of y=x^3/square root(x^4+1). limits of integration: x=0 and x=1

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to find the definite integral of y=x^3/square root(x^4+1) between the limits of integration: x=0 and x=1

let t = x^4 + 1

dt / 4 = x^3 dx

Int [x^3/sqrt(x^4+1) dx]

=> Int [ (1/4)/ sqrt t dt]

=> (1/4)*t^(1/2)/(1/2)

=> (1/2)sqrt t

substitute t = x^4 + 1

=> (1/2) sqrt (x^4 + 1)

Between the limits x = 0 and x = 1

(1/2) sqrt (1^4 + 1) - (1/2) sqrt ( 0 + 1)

=> (1/2) (sqrt 2 - 1)

The required definite integral is (1/2) (sqrt 2 - 1)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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^3dx/sqrt(x^4 + 1)

We notice that if we'll put x^4 + 1 = t and we'll differentiate, we'll get the numerator.

4x^3dx = dt

x^3dx = dt/4

We'll re-write the integral:

Int x^3dx/sqrt(x^4 + 1) = Int (dt/4)/sqrt t

Int (dt/4)/sqrt t = (1/4)*Int dt/sqrt t

(1/4)*Int dt/sqrt t = (1/4)*[t^(-1/2 + 1)/(-1/2 + 1)] + C

(1/4)*Int dt/sqrt t = (1/2)*sqrt t + C

Int f(x)dx = (1/2)*sqrt (x^4 + 1) + C

We'll evaluate the definite integral:

Int f(x)dx = (1/2)*sqrt (1^4 + 1) - (1/2)*sqrt (0^4 + 1)

Int f(x)dx = (1/2)*(sqrt2 - 1)

The definite integral of the function f(x) = x^3/sqrt(x^4 + 1), is Int f(x)dx = (1/2)*(sqrt2 - 1).

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