Hey guys I’m having trouble proving the area under the curve is the same as the arc length the question is:

f(x)=14ex+e−xf(x)= \frac{1}{4}e^x+e^{-x} , prove that’s the arc length on any interval has the same value as area under the curve.

For the arc length I got 14(ex−4e−x)+C\frac{1}{4}(e^x-4e^{-x})+C. But I didn’t get that for the area under the curve. Please help thank you !!

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For the area under the curve….you just integrate the e-powers…..Give that a try first.

– imranfat

Oct 21 at 2:13

I did and I just got 1/4(e^x)-e^-x

– Maiya Robinson

Oct 21 at 2:15

Well, take a look at Deepak’s suggestion.Once you have done your derivative work, it really boils done to an equation…

– imranfat

Oct 21 at 2:19

Okay thank you !

– Maiya Robinson

Oct 21 at 2:25

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1 Answer

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Hints:

Let y=f(x)\displaystyle y = f(x)

Arc length between X1X_1 and X2X_2 is given by ∫X2X1√1+(y′)2dx\displaystyle \int_{X_1}^{X_2} \sqrt{1 + (y’)^2}dx

Area under the curve between X1X_1 and X2X_2 is given by ∫X2X1ydx\displaystyle \int_{X_1}^{X_2} ydx

So all you need to do is prove that √1+(y′)2=y\displaystyle \sqrt{1 + (y’)^2} = y holds identically true.

Can you show that 1+(y′)2=y2\displaystyle {1 + (y’)^2} = y^2 for all xx?

+1) Hi there, where’s your cat?

– imranfat

Oct 21 at 2:20

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@imranfat It got transformed into a car. 🙂 Well, a car logo haha.

– Deepak

Oct 21 at 2:20

Okay I showed they equal , did you just square both sides to get ride of the square root on the left side and to make the y y^2 ?

– Maiya Robinson

Oct 21 at 2:24

@MaiyaRobinson Yes

– Deepak

Oct 21 at 2:28

Okay thank you so much !

– Maiya Robinson

Oct 21 at 2:30