Q.33 - Calculus

Question

For each function, $f$ and interval $[a,b]$, use definite integrals and the Fundamental Theorem of Calculus to find the exact values of (a) the signed area and (b) the absolute area of the region between the graph of  $f$and the x-axis from $x=a$ to $x=b$.

$f\left(x\right)={(2x-1)}^2-4, [a,b]=[-2,4]$.

Step-by-Step Solution

Answer

(a) The signed area is $54$.

(b) The absolute area is $\frac{194}{3}$.

Step 1. Given Information.

The function is,

$f\left(x\right)={(2x-1)}^2-4$.

The interval is, $[-2,4]$.

Part (a). The signed area.

The signed area is,

$$\begin{gathered} {\int }_{-2}^4{(2x-1)}^2-4dx={\int }_{-2}^{4}4x^2-4x-3dx \\ =4{\int }_{-2}^4{x}^{2}dx-4{\int }_{-2}^{4}xdx-3{\int }_{-2}^{4}dx \\ =4{\left[\frac{x^3}{3}\right]}_{-2}^4-4{\left[\frac{x^2}{2}\right]}_{-2}^4-3{\left[x\right]}_{-2}^4=4\left[\frac{64+8}{3}\right]-4\left[\frac{16-4}{2}\right]-3\left[6\right] \\ =96-24-18 \\ =54 \end{gathered}$$

Therefore, the signed area is $54$.

Part (b). The absolute area.

The graph of the function is,

The absolute area is,

$$\begin{gathered} {\int }_{-2}^4\left|{(2x-1)}^2-4\right|dx \\ ={\int }_{-2}^{-0.5}{(2x-1)}^2-4dx-{\int }_{-0.5}^{1.5}{(2x-1)}^2-4dx+{\int }_{1.5}^4{(2x-1)}^2-4dx={\int }_{-2}^{-0.5}(4x^2-4x-3)dx-{\int }_{-0.5}^{1.5}(4x^2-4x-3)dx+{\int }_{1.5}^4(4x^2-4x-3)dx \\ =\left[4\right[\frac{x^3}{3}]-4[\frac{x^2}{2}]-3{\left[x\right]]}_{-2}^{-0.5}-[4\left[\frac{x^3}{3}\right]-4\left[\frac{x^2}{2}\right]-3{\left[x\right]]}_{-0.5}^{1.5}+\left[4\right[\frac{x^3}{3}]-4[\frac{x^2}{2}]-3{\left[x\right]]}_{1.5}^4=\frac{27}{2}+\frac{16}{3}+\frac{275}{6} \\ =\frac{388}{6} \\ =\frac{194}{3} \end{gathered}$$