Question

$$ \begin{aligned} &\text { Describe why }\\\ &\int x\left(5-x^{2}\right)^{3} d x \neq \int u^{3} d u \text { where } u=5-x^{2} \end{aligned} $$

Short answer

The two integrals, \( \int x(5-x^{2})^{3} d x \) and \( \int u^{3} d u \), are not equal because simple substitution of \( u = 5 - x^2 \) into the integral does not account for the variable \( x \) in \( x dx \) term in the first integral. Therefore, the correct substitution would be recognizing that \( du = -2x dx \), and substitution should be as such to obtain \( -\frac{1}{2} \int u^3 du \), which leads to a different result when integrated.

Step-by-step solution

Describe the Integral with Substitution

Let's compute the integral \( \int u^{3} du \) with \( u = 5 - x^2 \). This is a simple case of integration and can be solved directly using the power rule for integrals, which states \(\int x^n dx = \frac{1}{n+1}x^{n+1} + C\), where \(C\) is the constant of integration. Here, \(n = 3\), so our integral will be \( \int u^{3} du = \frac{1}{3+1} u^{4} + C = \frac{1}{4} u^{4} + C \).

Explain Substitution

In the case of \( \int x(5 - x^2)^3 dx \), doing a direct substitution of \( u = 5 - x^2 \) would leave \( x \) as an extraneous variable in the integral. The integral would still contain \(x\) which cannot be expressed as a function of \( u \). This is not simply a case of substituting \(u\) into the integral.

Show the Correct Way to Approach the Integral

To correctly approach \( \int x(5 - x^2)^3 dx \), we would perform substitution by letting \(u = 5 - x^2\), such that \( du = -2x dx \), and \( -\frac{1}{2}du = x dx \). Substituting, the new integral is \( -\frac{1}{2} \int u^3 du\), which is different from the integral \( \int u^3 du \). The result obtained by integrating is different because we've accounted properly for the \( x dx \) term in the original integral.

Final Calculation

Now calculate the new integral, \( -\frac{1}{2} \int u^3 du = -\frac{1}{2} \frac{1}{4} u^{4} + C = -\frac{1}{8} u^{4} + C \). Substituting \( u = 5 - x^2 \) back in to obtain the final result.