1. ## Area Under Curve

$x=2-t^2$, $y= t(t+1)$

Find the exact area of the region bounded by the curve and the coordinate axes.

2. Originally Posted by Air
$x=2-t^2$, $y= t(t+1)$

Find the exact area of the region bounded by the curve and the coordinate axes.

To find the x-intercepts we set y =0 and solve for t.

$0=t(t+1) \\\ t=0 \\\ t=-1$

The area under the curve can be found using

$\int_{x_0}^{x_1}ydx$

Now we need to calculate dx so

$dx=-2t dt$

so we get

$\int_{1}^{2}ydx=\int_{-1}^{0}-2t^3-2t^2dt=-\frac{1}{6}$

Note that the area is negative because it is below the x-axis

Here is a graph. Good luck.

3. Now that I have read your question again, did you mean the part of the graph below the curve and inbetween the x and y axis. If so then

$\int_{0}^{1}ydx=\int_{-\sqrt{2}}^{-1}-2t^3-2t^2dt$

4. I don't know. The answer in mark scheme is...

$
2 + \frac43 \sqrt2
$

5. Okay this shaded part is the area they want...

$\int_{0}^{2}ydx=-\int_{2}^{0}ydx=-\int_{0}^{\sqrt{2}}t(t+1)(-2t)dt=\int_{0}^{\sqrt{2}}2t^3+2t^2dt=2+\frac{4}{3} \sqrt{2}$