Q. 40 - Calculus

Question

For each pair of functions f and g and each interval [a, b] in Exercises 38–40, use at least eight rectangles to approximate the area of the region between the graphs of f and g on the interval [a, b]. Your work should include graphs of f and g together with the rectangles that you used.

$f\left(x\right)=e^x,g\left(x\right)=\ln x,[0.5,2.5]$

Step-by-Step Solution

Answer

Area=9.89

Step 1. Given information

$f\left(x\right)=e^x,g\left(x\right)=\ln x,[0.5,2.5]$

Step 2. Area between the curve

$\begin{array}{r}{\int }_{0.5}^{2.5} \mid f\left(x\right)-g\left(x\right)dx={\int }_{0.5}^{0.75} \left(f\right(x)-g(x\left)\right)dx+{\int }_{0.75}^1 \left(f\right(x)-g(x\left)\right)dx+{\int }_1^{1.25} \left(f\right(x)-g(x\left)\right)dx\\ +{\int }_{1.25}^{1.5} \left(f\right(x)-g(x\left)\right)dx+{\int }_{1.5}^{1.75} \left(f\right(x)-g(x\left)\right)dx+{\int }_{1.75}^{2.0} \left(f\right(x)-g(x\left)\right)dx\\ +{\int }_{2.0}^{2.25} \left(f\right(x)-g(x\left)\right)dx+{\int }_{2.25}^{2.5} \left(f\right(x)-g(x\left)\right)dx\\ {\int }_{0.5}^{25} \mid f\left(x\right)-g\left(x\right)dx={\int }_{0.5}^{0.75} \left(e^x-\ln \left(x\right)\right)dx+{\int }_{0.75}^1 \left(e^x-\ln \left(x\right)\right)dx+{\int }_1^{1.25} \left(e^x-\ln \left(x\right)\right)dx\\ +{\int }_{1.25}^{1.5} \left(e^x-\ln \left(x\right)\right)dx+{\int }_{1.5}^{1.75} \left(e^x-\ln \left(x\right)\right)dx+{\int }_{1.75}^{2.0} \left(e^x-\ln \left(x\right)\right)dx\\ +{\int }_{20}^{2.25} \left(e^x-\ln \left(x\right)\right)dx+{\int }_{2.25}^{2.5} \left(e^x-\ln \left(x\right)\right)dx\end{array}$

$\begin{array}{r}={\left[e^x-x\ln x+x\right]}_{0.5}^{0.75}+{\left[e^x-x\ln x+x\right]}_{0.75}^1+{\left[e^x-x\ln x+x\right]}_1^{1.25}\\ +{\left[e^x-x\ln x+x\right]}_{1.25}^{1.5}+{\left[e^x-x\ln x+x\right]}_{1.5}^{1.75}+{\left[e^x-x\ln x+x\right]}_{1.75}^{2.0}\\ +{\left[e^x-x\ln x+x\right]}_{2.0}^{2.25}+{\left[e^x-x\ln x+x\right]}_{2.25}^{2.5}\\ =0.587466+0.63552+0.743132+0.91208+1.15178\\ +1.47748+1.91038+2.4786\\ =9.896438\end{array}$