Question

To evaluate the integral function \(\int_0^1 x \sqrt {1 - {x^4}} dx\).

Short answer

The evaluation of the integral is \(\frac{1}{8}\pi \).

Step-by-step solution

Given data

The integral function is \(\int_0^1 x \sqrt {1 - {x^4}} dx\).

The region lies between \(x = 0\) and \(x = 1\).

The substitution is \(u = {x^2}\)

Concept used of Substitution rule

The Substitution Rule:

If\(u = g(x)\)is a differentiable function whose range is an interval\(l\)and\(f\)is continuous on\(l\), then\(\int f (g(x)){g^\prime }(x)dx = \int f (u)du.{\rm{ }}\)

Simplify the function

Consider

\(u = {x^2}\) ….. (1)

Differentiate both sides of the Equation (1).

\(\begin{aligned}{l}du = 2xdx\\\frac{1}{2}du = xdx\end{aligned}\)

Calculate the lower limit value of \(u\) by the use of Equation (1).

Substitute 0 for \(x\) in Equation (1).

\(u = {0^2} = 0\)

Calculate the upper limit value of \(u\) by the use of Equation (1).

Substitute 1 for \(x\) in Equation (1).

\(u = {1^2} = 1\)

The integral function is,

\(\int_0^1 x \sqrt {1 - {x^4}} dx\) …..(2)

Apply lower and upper limits

Apply lower and upper limits for \(u\) in Equation (2).

Substitute \(u\) for \(\left( {{x^2}} \right)\) and \(\left( {\frac{1}{2}du} \right)\) for \((xdx)\) in Equation (2).

\(\begin{aligned}{c}\int_0^1 x \sqrt {1 - {x^4}} dx = \int_0^1 {\sqrt {1 - {u^2}} } \left( {\frac{1}{2}du} \right)\\ = \frac{1}{2}\int_0^1 {\sqrt {1 - {u^2}} } du\end{aligned}\)

Interpret the integral function.as the area of a quarter-circle with radius 1.

The integral function \((I)\) equals to the area of a quarter-circle with radius 1.

Modify Equation (3) as follows