Q. 18 - Calculus

Question

Consider the integral $\int \frac{x^{-2}-4}{x^3}dx$.

(a) Solve this integral by using u-substitution.

(b) Solve the integral another way, using algebra to simplify the integrand first.

(c) How must your two answers be related? Use algebra to prove this relationship.

Step-by-Step Solution

Answer

(a) The value of integral by using u-substitution $\int \frac{x^{-2}-4}{x^3}dx=-\frac{{(x^{-2}-4)}^2}{4}+C$

(b) The the value integral using algebra to multiply out the integrand first is $\int \frac{x^{-2}-4}{x^3}dx=-\frac{1}{4x^4}+\frac{2}{x^2}+C$.

(c) The value of both answer differ by a constant $-\frac{{(x^{-2}-4)}^2}{4}=-\frac{1}{4x^4}+\frac{2}{x^2}-4$.

Step 1. Given Information

Consider the integral $\int \frac{x^{-2}-4}{x^3}dx$.

(a) Solve this integral by using u-substitution.

(b) Solve the integral another way, using algebra to simplify the integrand first.

(c) How must your two answers be related? Use algebra to prove this relationship.

Part (a) Step 1. Solve this integral by using u-substitution.

Let 

$$\begin{gathered} u=x^{-2}-4 \\ \frac{du}{dx}=-2x^{-3} \\ \frac{du}{dx}=-2\frac{1}{x^3} \\ -\frac{1}{2}du=\frac{1}{x^3}dx \end{gathered}$$

Part (a) Step 2. This substitution changes the integral into

$$\begin{gathered} \int \frac{x^{-2}-4}{x^3}dx=-\frac{1}{2}\int udu \\ \int \frac{x^{-2}-4}{x^3}dx=-\frac{1}{2}\left[\frac{u^{1+1}}{1+1}\right]+C \\ \int \frac{x^{-2}-4}{x^3}dx=-\frac{1}{2}\left[\frac{u^2}{2}\right]+C \\ \int \frac{x^{-2}-4}{x^3}dx=-\frac{u^2}{4}+C \\ \int \frac{x^{-2}-4}{x^3}dx=-\frac{{(x^{-2}-4)}^2}{4}+C \end{gathered}$$

Part (b) Step 1. Solving integral using algebra to multiply out the integrand first.

$$\begin{gathered} \int \frac{x^{-2}-4}{x^3}dx=\int \frac{1}{x^3}\left(x^{-2}-4\right)dx \\ \int \frac{x^{-2}-4}{x^3}dx=\int \frac{1}{x^3}\left(\frac{1}{x^2}-4\right)dx \\ \int \frac{x^{-2}-4}{x^3}dx=\int \left(\frac{1}{x^3}·\frac{1}{x^2}-\frac{1}{x^3}·4\right)dx \\ \int \frac{x^{-2}-4}{x^3}dx=\int \left(\frac{1}{x^5}-\frac{4}{x^3}\right)dx \end{gathered}$$

∫ x − 2 − 4 x 3 d x = ∫ 1 x 5 − 4 x 3 d x

$$\begin{gathered} \int \frac{x^{-2}-4}{x^3}dx=\int \frac{1}{x^5}dx-\int \frac{4}{x^3}dx \\ \int \frac{x^{-2}-4}{x^3}dx=\int x^{-5}dx-4\int x^{-3}dx \\ \int \frac{x^{-2}-4}{x^3}dx=\left[\frac{x^{-5+1}}{-5+1}\right]-4\left[\frac{x^{-3+1}}{-3+1}\right]+C \\ \int \frac{x^{-2}-4}{x^3}dx=\left[\frac{x^{-4}}{-4}\right]-4\left[\frac{x^{-2}}{-2}\right]+C \\ \int \frac{x^{-2}-4}{x^3}dx=-\frac{1}{4x^4}+\frac{2}{x^2}+C \end{gathered}$$

Part (c) Step 1. Using algebra to prove this relationship.

The value of integral in part (a) is  

$\int \frac{x^{-2}-4}{x^3}dx=-\frac{{(x^{-2}-4)}^2}{4}+C$

The value of integral in part (b) is  

$\int \frac{x^{-2}-4}{x^3}dx=-\frac{1}{4x^4}+\frac{1}{2x^2}+C$

The value of both answer differ by a constant 

$$\begin{gathered} -\frac{{(x^{-2}-4)}^2}{4}=-\left(\frac{{\left(x^{-2}\right)}^2-2\times x^{-2}\times 4+{\left(4\right)}^2}{4}\right) \\ -\frac{{(x^{-2}-4)}^2}{4}=\frac{-x^{-4}+8x^{-2}-16}{4} \\ -\frac{{(x^{-2}-4)}^2}{4}=\frac{-x^{-4}}{4}+\frac{8x^{-2}}{4}-\frac{16}{4} \\ -\frac{{(x^{-2}-4)}^2}{4}=\frac{-x^{-4}}{4}+2x^{-2}-4 \\ -\frac{{(x^{-2}-4)}^2}{4}=-\frac{1}{4x^4}+\frac{2}{x^2}-4 \end{gathered}$$