Question

If \(f\) is continuous and \(\int_0^9 f (x)dx = 4\), find \(\int_0^3 x f\left( {{x^2}} \right)dx\).

Short answer

The value of the integral is: \(\int\limits_0^3 {xf({x^2})dx = 2} \)

Step-by-step solution

Step 1: Given information

If\(f(x)\)is a function defined on an interval\((a,b)\), the definite integral of\(f\)from\(a\)to\(b\)is given by

\(\int_a^b f (x)dx = \mathop {\lim }\limits_{n \to \infty i = 1}^n f\left( {x_i^*} \right)x{\rm{, }}\)

provided the limit exists. If this limit exists, the function \(f(x)\)is said to be integrable on \((a,b)\), or is an integrable function.

To find the value of integral

To Find the function

\(\int\limits_0^3 {xf({x^2})dx} \)

Substitute \(u = {x^2}\) into the expression:

\(\int\limits_0^3 {xf(u)dx} \)

To find the value of the integral, first determine\(du\)by finding the derivative of\(u = {x^2}\):

\(du = {({x^2})^'}\)

Calculate the right-side equation 

Calculate the derivation of the right side of the equation:

\(du = 2xdx\)

Divide both sides of the equation by\(2\):

\(\frac{{du}}{2} = \frac{{2xdx}}{2}\)

\(\frac{{du}}{2} = \frac{{\cancel{2}xdx}}{{\cancel{2}}}\)

Reduce the fraction:

\(\frac{{du}}{2} = xdx\)

Swap the sides of the equation:

\(xdx = \frac{{du}}{2}\)

Since the lower limit of the integral is\(0\)and\(u = {x^2}\), it follows that the lower limit of the new integral is:

\(u = 0 \times 0 = 0\)

Since the upper limit of the integral is\(3\)and\(u = 2{x^2}\), it follows that the lower limit of the new integral is:

\(u = 3 \times 3 = 9\)

\(\int\limits_0^3 {f(u)xdx} \)

Substitute the lower and upper limits

Substitute lower limit with\(0\), upper limit with\(9\), and\(xdx = \frac{{du}}{2}\)into the expression:

\(\int\limits_0^9 {f(u)\frac{{du}}{2}} \)

Rewrite the expression:

\(\int\limits_0^9 {\frac{1}{2}} f(u)du\)

Final proof

By using the constant property of integrals, it follows:

\(\frac{1}{2}\int\limits_0^9 {f(u)du} \)

Since\(\int\limits_0^9 {f(x)dx = 4} \), it follows:

\(\frac{1}{2} \times 4 = 2\)

Hence, the value of the integral is: \(\int\limits_0^3 {xf({x^2})dx = 2} \)