I antiderived and got x-x^3/3 then subbed the 0 and 2's in but my final answer was totally wrong .Can anyone help with the process etc ...the final answer is 2 units squared but I seem to get 8/3 please help ?
Thanks
I antiderived and got x-x^3/3 then subbed the 0 and 2's in but my final answer was totally wrong .Can anyone help with the process etc ...the final answer is 2 units squared but I seem to get 8/3 please help ?
Thanks
From x = 0 to x = 1, the area is above the x-axis.
From x = 1 to x = 2, the area is below the x-axis
$\displaystyle A = \int_0^1 1-x^2 \, dx - \int_1^2 1-x^2 \, dx$
try again ...
Thanks skeeter ,
how do I know when it is above the x axis or below the x axis when I look at the equation any advice would be greatly appreciated ,as I don't really know when to apply that rule.
cheers
$1-x^2=0\Longrightarrow x=\pm 1$
This tells you when it touches the $x$-axis. For a continuous function (like a polynomial), it will never "jump" from positive to negative. It will always go through the $x$-axis. So, at any value $x>1$ you find the sign of $1-x^2$, it will have the same sign for all $x>1$. Same for $-1 <x <1$.
Have you completed a course in precalculus?
Transformations of basic parent functions and the sketching of their graphs is one of the major topics covered.
I would expect my calculus students to know that the graph of $y=1-x^2$ is an inverted parabola with vertex at (0,1) and zeros at x = -1 and x = 1.
thanks guys ,I will study the graphs ,I have done graph sketching and as this course I'm doing goes quite quickly I tend to forget things learnt earlier ...really appreciate the help.
I have calculated the answer to be 2 and 2/3 the area under the curve from my calculation was -2 and the area above was 2/3 the answer from the work sheet is 2 ..can u guys tell me what i did wrong or if the worksheet is wrong cheers
Thanks skeeter ,
I did x-x^3/3 - 0-0/3 for the first part being 1 - 1/3 which cam to be 2/3 and the second part 2-8/3-1-1/3 for the second part of the integral being -2
if the answer is 2 I've obviously subbed in the wrong numbers or did something wrong .
$\displaystyle A = \int_0^1 1-x^2 \, dx - \int_1^2 1-x^2 \, dx$
$\bigg[x-\dfrac{x^3}{3}\bigg]_0^1 - \bigg[x - \dfrac{x^3}{3}\bigg]_1^2$
$\bigg[\left(1-\dfrac{1}{3}\right) -0\bigg] - \bigg[\left(2-\dfrac{8}{3}\right) - \left(1-\dfrac{1}{3}\right)\bigg]$
$\bigg[\dfrac{2}{3}\bigg] - \bigg[\left(-\dfrac{2}{3}\right) - \left(\dfrac{2}{3}\right)\bigg]$
$\dfrac{2}{3} - \left(-\dfrac{4}{3}\right) = \dfrac{2}{3} + \dfrac{4}{3} = \dfrac{6}{3} = 2$
my god! i think i truly need to be labelled a retard simple algebra did it to me ...the minus outside the brackets and multiplying to make a positive was my mistake ...thanks again skeeter ,you are the true math master of all math ..
cheers