4.1.237 AP calculus exam find area

[karush]

$\tiny{237}$
$\textsf{The total area of the region bounded by the graph of $f(x)=x\sqrt{1-x^2}$ and the x-axis is}$
$$(A) \dfrac{1}{3}\quad
(B) \dfrac{1}{3}\sqrt{2}\quad
(C) \dfrac{1}{2}\quad
(D) \dfrac{2}{3}\quad
(E) 1 $$
find the limits of integration if
$$f(x)=0 \textit{ then } x=-1,0,1$$
so we have a symetric graph about $y=x$ so
$$\displaystyle
2\left|\int_{0}^1 x\sqrt{1-x^2}\,dx\right|
=2\left(
-\frac{1}{3}\left(1-x^2\right)^{\frac{3}{2}}
\right)\bigg|_{0}^1
=\dfrac{2}{3}\quad (D)$$ok hopefully this is the answer but I was alarmed how much time I spent on calculation
which will kill you on these exams even if you get the corrects answers.not real sure what the slam dunk method would be just from obersavation.

 

[topsquark]

$\tiny{237}$
$\textsf{The total area of the region bounded by the graph of $f(x)=x\sqrt{1-x^2}$ and the x-axis is}$
$$(A) \dfrac{1}{3}\quad
(B) \dfrac{1}{3}\sqrt{2}\quad
(C) \dfrac{1}{2}\quad
(D) \dfrac{2}{3}\quad
(E) 1 $$
find the limits of integration if
$$f(x)=0 \textit{ then } x=-1,0,1$$
so we have a symetric graph about $y=x$ so
$$\displaystyle
2\left|\int_{0}^1 x\sqrt{1-x^2}\,dx\right|
=2\left(
-\frac{1}{3}\left(1-x^2\right)^{\frac{3}{2}}
\right)\bigg|_{0}^1
=\dfrac{2}{3}\quad (D)$$ok hopefully this is the answer but I was alarmed how much time I spent on calculation
which will kill you on these exams even if you get the corrects answers.not real sure what the slam dunk method would be just from obersavation.

Looks good to me, although since you didn't mention what method you used to integrate. A simple substitution [tex]u = 1 - x^2[/tex] looks doable.

Also, this graph is not symmetric. It is antisymmetric.

-Dan

 

[karush]

ok its not mirrored
I think some have called it symeteric about origin?

the integral is also a table reference but nobody could remember all of those

Mahalo

 

[HallsofIvy]

"From observation" you should immediately see the "[tex]1- x^2[/tex]" inside the square root and the "x" outside the root and think "Aha, I can let [tex]u= 1- x^2[/tex[ so that [tex]dx= -2xdx[/tex] and I have the 'x' to give me 'xdx'!"

 

[karush]

"From observation" you should immediately see the "[tex]1- x^2[/tex]" inside the square root and the "x" outside the root and think "Aha, I can let [tex]u= 1- x^2[/tex[ so that [tex]dx= -2xdx[/tex] and I have the 'x' to give me 'xdx'!"

total awesome...