2 times the integral (from -3 to 3) of (4-x) (sqrt(9-x^2) ) dx.
Thank you very much.
$\displaystyle
\int_{-3}^3 (4-x) \sqrt{9-x^2}~dx = \int_{-3}^3 4\sqrt{9-x^2}~dx - \int_{-3}^3 x \sqrt{9-x^2}~dx
$
and these two integrals can be done by the usual book methods, the first by a trig substitution and the second
by looking at the derivative of $\displaystyle (9-x^2)^{3/2}$.
RonL
$\displaystyle
\int_{-3}^3 (4-x) \sqrt{9-x^2}~dx
$
Hint:
$\displaystyle = 4 \int_{-3}^3 \sqrt{9-x^2}~dx - \int_{-3}^3 x\sqrt{9-x^2}~dx$
You can do the second integral with a simple substitution. So let's look at the first term:
$\displaystyle \int_{-3}^3 \sqrt{9-x^2}~dx$
Let $\displaystyle x = 3y \implies dx = 3 dy$
Thus
$\displaystyle \int_{-3}^3 \sqrt{9-x^2}~dx = \int_{-1}^1 \sqrt{9-(3y)^2}~3dy$
$\displaystyle = 3\int_{-1}^1 \sqrt{9-9y^2}~dy$
$\displaystyle = 3\int_{-1}^1 3\sqrt{1-y^2}~dy$
$\displaystyle = 9\int_{-1}^1 \sqrt{1-y^2}~dy$
Does this integral look familiar?
-Dan
Hi Dan,
how do I end this one?
9 times the integral from -1 to 1 of sqrt(1-y^2) dy - your last step! Really don't know how
Also,
Can anyone remind me how to do the trig substitution? I forgot it ! Please show me the steps . Thanks.
Change of variables according to $\displaystyle y=\sin u$ implies
$\displaystyle \begin{aligned}
\int {\sqrt {1 - y^2 } \,dy} &= \int {\sqrt {1 - \sin ^2 u} \cos u\,du}\\
&= \int {\cos u\cos u\,du}\\
&= \int {\cos ^2 u\,du}
\end{aligned}$
Which is easy to take.