Q. 56 - 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.)

${\int }_{-\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{x}^2\left(se{c}^2\right(x^2\left)\right)dx$

Step-by-Step Solution

Answer

${\int }_{-\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{x}^2\left(se{c}^2\right(x^2\left)\right)dx=\frac{2}{3}\tan \frac{{\mathrm{\pi }}^3}{64}$.

Step 1. Given information.

A definite integral is given as ${\int }_{-\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{x}^2\left(se{c}^2\right(x^2\left)\right)dx$.

Step 2. Using the Fundamental theorem of Calculus.

Let

$$\begin{gathered} f\left(x\right)=\tan x \\ g\left(x\right)=x^3 \end{gathered}$$

such that

$$\begin{gathered} f\text{'}\left(x\right)=se{c}^{2}x \\ g\text{'}\left(x\right)=3x^2 \\ f\text{'}\left(g\right(x\left)\right)=se{c}^2\left(x^3\right) \end{gathered}$$

Now we get

$$\begin{gathered} {\int }_{-\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{x}^2\left(se{c}^2\right(x^2\left)\right)dx \\ =\frac{1}{3}{\int }_{-\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}3x^2\left(se{c}^2\right(x^2\left)\right)dx \\ =\frac{1}{3}{\left[\tan \right(x^3\left)\right]}_{-\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.} [f\text{'}(g\left(x\right))g\text{'}(x)dx=f(g\left(x\right)\left)\right] \\ =\frac{1}{3}\left[\tan \right(\frac{{\mathrm{\pi }}^3}{64})-\tan (-\frac{{\mathrm{\pi }}^3}{64}\left)\right] \\ =\frac{2}{3}\tan \frac{{\mathrm{\pi }}^3}{64} \end{gathered}$$

The exact value of the given definite integral is $\frac{2}{3}\tan \frac{{\mathrm{\pi }}^3}{64}$.

Step 3. The graph to verify the answer is

The solution is area under graph which is 

$$\begin{gathered} a\approx 0.385747 \\ \approx \frac{2}{3}\tan \frac{{\mathrm{\pi }}^3}{64} \end{gathered}$$