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Area between a curve and a tangent!

The tangent to the curve y=5+4x-x² at A(3,8) meets the x-axis at B.
find the shaded area.

so what i did is.... i found the tangent equation which is y=-2x+14

and then i integrated from 3 to 7 for y=-2x+14 - from 3 to 7 for y=5+4x-x²to give a final answer... 10.7

However, i was wondering would it work the same way if i took the tangent equation to be y=-2x+8 ??

either way? is there's one right and one wrong?
if so.. what is the tangent equation that i should use?

Reply 1

deema21

just to clarify what i meant

\int^7_3 -2x+14\ dx

\int^7_3 5+4x-x^2\ dx

\int^7_3 -2x+8\ dx

\int^7_3 5+4x-x^2\ dx

Reply 2

iknowi2

Original post by deema21

just to clarify what i meant

\int^7_3 -2x+14\ dx

\int^7_3 5+4x-x^2\ dx

\int^7_3 -2x+8\ dx

\int^7_3 5+4x-x^2\ dx

No, they are two different expressions; (-2x + 14) does not equal (-2x + 8). I'm assuming (-2x + 8) is from the markscheme, and therefore your own version which is (-2x + 14) is incorrect. If so, this is simply a teething error, check your linear equation again.

Reply 3

deema21

Original post by Konnichiwa

All you have to do is find the integral between x = 3 and x = 7.

Then, find the area of the triangle between x = 3, x = 7, with a height (you can find this with the y coordinate via simultaneous equations), 1/2(base * height).

Then subtract the area of the triangle by the integral that you just calculated.

that's fair enough!!!
thanks! :smile:

Reply 4

steve2005

Original post by deema21

to give a final answer... 10.7

I get area to be 20/3 and not 10.7

Reply 5

ghostwalker

Original post by deema21

...

I agree with steve2005; area is 20/3.

The area under your curve is found by integrating between 3 and 5, not 3 and 7, and this is subtracted from the area of the triangle.