3
$\begingroup$

Find the area of the region bounded by the parabola $y = 4x^2$, the tangent line to this parabola at $(2, 16)$, and the $x$-axis.

I found the tangent line to be $y=16x-16$ and set up the integral from $0$ to $2$ of $4x^2-16x+16$ with respect to $x$, which is the top function when looking at the graph minus the bottom function. I took the integral and came up with $\frac{4}{3}x^3-8x^2+16x$ evaluated between $0$ and $2$. This came out to be $\frac{32}{3}$ but this was the incorrect answer. Can anyone tell me where I went wrong?

$\endgroup$
0
$\begingroup$

It isn't right. Try to draw such region.

Hint: After drawing it, note that you have to calculate $\int_0^1 4x^2\;dx + \int_1^2 4x^2-16x+16\;dx$.

$\endgroup$
  • $\begingroup$ I did this and got 12, but this was still incorrect. Any suggestions? $\endgroup$ – Maggie Nov 22 '16 at 0:14
  • $\begingroup$ I got $\frac{8}{3}$. I'm sorry but did you do it right? $\endgroup$ – Rodrigo Dias Nov 22 '16 at 0:22
  • $\begingroup$ Oh I see what I did. I now got 8/3 as well. Thanks! $\endgroup$ – Maggie Nov 22 '16 at 0:25
  • $\begingroup$ Any time! ${}{}$ $\endgroup$ – Rodrigo Dias Nov 22 '16 at 0:28
0
$\begingroup$

The tangent crosses the $x$ axis at $x=1$, so your integral is including (with the plus sign) also the triangle made by the tangent below the $x$ axis.
The correct way is to integrate only the parabola for $x=0 \cdots 2$ (which is $32/3$ and then subtract the area of the triangle$(1,0),(2,16),(2,0)$, which is $8$, so the net area is $8/3$ .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.