Question

\(\int\limits_{ - 1}^1 {({x^5} - 6{x^9} + \frac{{sinx}}{{{{(1 + {x^4})}^2}}})dx} \)

Short answer

The answer is TRUE.

Step-by-step solution

Step 1: Use of Theorem on Integrals of Symmetric Function

Let us use Theorem on integrals of symmetric functions. We get

\(f(x) = {x^5} - 6{x^9} + \frac{{\sin x}}{{{{(1 + {x^4})}^2}}}\)

Step 2: Find the value of f(-x)

\(\begin{array}{l}f( - x) = {( - x)^5} - 6{( - x)^9} + \frac{{\sin ( - x)}}{{{{(1 + {{( - x)}^4})}^2}}}\\ = - {x^5} + 6{x^9} - \frac{{\sin (x)}}{{{{(1 + {x^4})}^2}}}\end{array}\)

\( = - \left( {{x^5} - 6{x^9} + \frac{{\sin (x)}}{{{{(1 + {x^4})}^2}}}} \right)\)

\( = - f(x)\)

Step 3: Use of Theorem on Integrals of Symmetric Function

Hence f(X) is odd so using Theorem on integrals of Symmetric Functions

\(\int\limits_{ - 1}^1 {\left( {{x^5} - 6{x^9} + \frac{{\sin (x)}}{{{{(1 + {x^4})}^2}}}} \right)dx = 0} \)

So the given statement is True.