Question

calculus help!!!

Answer 1

\[\int\limits_{2}^{0}\left| 4-x ^{2} \right|dx \] I keep getting 0 as the answer, but my teacher tells me that is wrong... How do you do this then?

Answer 2

what is the topic name?

Answer 3

definite integrals?

Answer 4

4-x^2 dx=4x-1/3x^3
from 0 t0 2
[4(0)-1/3(0)^3]-[ 4(2)-1/3(2)^3]
|0-8+8/3|=|(-24+8)/3|=|-16/3|=16/3

Answer 5

Note that \(4-x^2 \geq0\) in between \(x=0\) and \(x=2\), so absolute value doesn't do anything here.

So \[\int\limits_{2}^{0}\left| 4-x ^{2} \right|dx = \int\limits_{2}^{0} 4-x ^{2} dx \]

So \[\int\limits_{2}^{0} 4-x ^{2}dx = \int\limits_{2}^{0} 4x^0dx-\int\limits_{2}^{0}x ^{2}dx = \left.4x-\dfrac{1}{3}x^3~\right|_{2}^0\]

Answer 6

I think you should divide the interval into parts when solving the | |
as we know
\[|4-x^2|=\begin{cases}4-x^2 ~~~~if~~~4-x^2>0~~~that~~~is~~~-2<x<2\\-(4-x^2)~~~if~~~ (4-x^2)<0~~~that~~ is~~ x<-2~~~or~~~x>2\end{cases}\]

Answer 7

Your integral is from 2 to 0 = - integral from 0 to 2,
That is int the first case, hence take
\[-\int_0^2 (4-x^2 )dx= -(4x-x^3/3)|^2_0\]