Question

\[A~curve~is~given~by~the~equation~ y={2 \over 3}e^x+ {1 \over 3}e^{-2x}.\](a)Evaluate a definite integral to find the area between the curve, the x-axis and the lines x=0 and x=1, showing your working

(b)Use calculus to determine whether the turning point at the point where x=0 is a maximum or minimum.

Answer 1

\[\int\limits_{}^{}(\frac{2}{3}e ^{x}+\frac{1}{3}e ^{-2x})dx\]

Answer 2

\[\huge [\frac{2}{3}e ^{x}+\frac{1}{3}\frac{1}{-2}e ^{-2x}]^{1}_{0}\]
\[\huge [\frac{2}{3}e ^{x}+\frac{1}{-6}e ^{-2x}]^{1}_{0}\]

Answer 3

For b), differentiate y
\[\frac{dy}{dx}=\frac{2}{3}e ^{x}-\frac{2}{3}e ^{-2x}\]

Answer 4

\[\frac{dy}{dx}=0\]
\[\frac{2}{3}e ^{x}-\frac{2}{3}e ^{-2x}=0\]
Solve for x

Answer 5

To find whether is a max or min,
\[\frac{d ^{2}y}{dx ^{2}}\]
If value of \[\frac{d ^{2}y}{dx ^{2}}>0\]
Then it is a minimum.
If value of
\[\frac{d ^{2}y}{dx ^{2}}<0\]
Then it is a maximum.

Answer 6

How do you differentiate it the second time?

Answer 7

Use the differentiated equation differentiate again

Answer 8

What will that give? \[{2 \over 3}e -{1\over12}e^{-2x}?\]

Answer 9

*2/3 e^x* sorry

Answer 10

\[\frac{d^{2}y}{dx ^{2}}=\frac{2}{3}e ^{x}+\frac{4}{3}e ^{-2x}\]

Answer 11

Why?

Answer 12

differentiate \[e ^{-2x}\]
\[\frac{dy}{dx}=-2e ^{-2x}\]