Question

opinion about method please. (To be posted)

(I posted there, but I like to know what y'all think)



(originally from)
http://openstudy.com/users/sithsandgiggles#/updates/54a9adc6e4b0e4319f90f72f

Answer 1

\(\large\color{slate}{ f(x)=e^{0.4x} }\) \(\LARGE\color{white}{ \left| \right| }\)
Point R \(\large\color{slate}{ (5,7.39) }\) \(\LARGE\color{white}{ \left| \right| }\)

Find the are under the curve, of f(x), bounded by y-axis, and the line tangent to f(x) at point R.

Answer 2

` The works done: `
\(\LARGE\color{white}{ \left| \right| }\)
the slope, \(\LARGE\color{white}{ \left| \right| }\)
\(\large\color{slate}{ f'(x)=(0.4)e^{0.4x} }\) \(\LARGE\color{white}{ \left| \right| }\)
\(\large\color{slate}{ f'(5)=(0.4)e^{0.4 \times (5)}=2.956 }\) \(\LARGE\color{white}{ \left| \right| }\)
\(\LARGE\color{white}{ \left| \right| }\)
the tangent line, \(\LARGE\color{white}{ \left| \right| }\)
\(\large\color{slate}{ y-7.39=2.956(x-5) }\) \(\LARGE\color{white}{ \left| \right| }\)
\(\large\color{slate}{ y=2.956(x-5) +7.39 }\) \(\LARGE\color{white}{ \left| \right| }\)
\(\large\color{slate}{ y=2.956x-7.39 }\) \(\LARGE\color{white}{ \left| \right| }\)
\(\LARGE\color{white}{ \left| \right| }\)
x-intercept. \(\LARGE\color{white}{ \left| \right| }\)
\(\large\color{slate}{ 0=2.956x-7.39 }\) \(\LARGE\color{white}{ \left| \right| }\)
\(\large\color{slate}{ 7.39=2.956x }\) \(\LARGE\color{white}{ \left| \right| }\)
\(\large\color{slate}{ 2.5=x }\) \(\LARGE\color{white}{ \left| \right| }\)
\(\LARGE\color{white}{ \left| \right| }\)
the \(\large\color{slate}{ \triangle }\): \(\LARGE\color{white}{ \left| \right| }\)
`(2.5,0)` , `(5,0)` , `(5,7.39)` \(\LARGE\color{white}{ \left| \right| }\)
\(\LARGE\color{white}{ \left| \right| }\)
area of the (above) \(\large\color{slate}{ \triangle }\)
( \(\large\color{black}{\color{red}{\rm length} \times \color{blue}{\rm height }}\) ) \(\LARGE\color{white}{ \left| \right| }\)
\(\large\color{black}{A_\triangle= (\color{red}{5-2.5}) \times \color{blue}{(7.39) }\div 2}\) \(\LARGE\color{white}{ \left| \right| }\)
\(\large\color{black}{A_\triangle= (\color{red}{2.5}) \times \color{blue}{(7.39) }\div 2}\) \(\LARGE\color{white}{ \left| \right| }\)
\(\large\color{black}{A_\triangle= 4.7375}\) \(\LARGE\color{white}{ \left| \right| }\)

The area of the region is given by:
\(\large\color{slate}{\displaystyle\int\limits_{0}^{5}f(x)~dx-A_\triangle}\)

final calculations:
\(\large\color{slate}{\displaystyle\int\limits_{0}^{5}e^{0.4x}~dx=~2.5e^{0.4x}~{{\Huge |}_{_{\Large 0}}^{^{\Large 5}}}~~~~~=18.4726-2.5=\color{blue}{15.9726}}\)

\(\large\color{slate}{15.9726-4.7375=11.2351}\)