Q. 54 - Calculus

Question

Use the Fundamental Theorem of Calculus to find the exact

values of each of the definite integrals in Exercises 19–64. Use

a graph to check your answer. (Hint: The integrands that involve

absolute values will have to be considered piecewise.)

${\int }_1^2\frac{x^2{e}^x-2x{e}^x}{x^4}dx$

Step-by-Step Solution

Answer

${\int }_1^2\frac{x^2{e}^x-2x{e}^x}{x^4}dx=\frac{e^2}{4}-e$.

Step 1. Given information.

A definite integral is given as ${\int }_1^2\frac{x^2{e}^x-2x{e}^x}{x^4}dx$.

Step 2. Using the Fundamental theorem of Calculus.

Let 

$$\begin{gathered} f\left(x\right)=e^x \\ g\left(x\right)=x^2 \end{gathered}$$

then

$$\begin{gathered} \frac{d}{dx}\frac{f\left(x\right)}{g\left(x\right)} \\ =\frac{f\text{'}\left(x\right)g\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{(g{\left(x\right))}^2} \\ =\frac{x^2{e}^x-2x{e}^x}{x^4} \end{gathered}$$

Now we have

$$\begin{gathered} {\int }_1^2\frac{x^2{e}^x-2x{e}^x}{x^4}dx \\ ={\left[\frac{e^x}{x^2}\right]}_1^2 \\ =[\frac{e^2}{2^2}-\frac{e}{1}] \\ =\frac{e^2}{4}-e \end{gathered}$$

The exact value of the given definite integral is $\frac{e^2}{4}-e$.

Step 3. The graph to verify the answer is

The solution is area under graph which is

$$\begin{gathered} a\approx -0.871017 \\ \approx \frac{e^2}{4}-e \end{gathered}$$