Areas under/between curves. (1 Viewer)

Psychosomatical · Aug 4, 2009

Just need an answer here, as i already have one, just need to check.

The graphs of y = -cosx and y = (pi^2)/4 - x^2 intersect on teh x-axis at x = pi/2 and x = -pi/2. Attached is a dodgy graph, to see if it sheds any light on the area i'm looking for.

thanks

Drongoski · Aug 4, 2009

Answer appears to be, taking into account symmetry about the y-axis:

$\100dpi Area\ =\ 2\times \int_{0}^{\frac{\pi }{2}}\ \left [ (\frac{\pi ^{2}}{4}-x^{2})\ -\ (-cos\ x) \right ]\ dx\\ \\ =\ \ 2\times \int_{0}^{\frac{\pi }{2}}\ (\frac{\pi ^{2}}{4}-x^{2}+cos\ x)\ dx\\ \\ =\ \ 2\times \left [ \frac{\pi^{2} }{4}x-\frac{x^{3}}{3}+sin\ x \right ]^{\frac{\pi }{2}}_{0}\ \ =\ \ 2\left ( \frac{\pi ^{2}}{4}\cdot \frac{\pi }{2}-\frac{1}{3}(\frac{\pi }{2})^{3}+sin\frac{\pi }{2} \right )\ -\ 2(0)\\ \\ =\ \ \frac{\pi ^{3}}{6}+2\ \ unit^{2}$

Psychosomatical · Aug 4, 2009

Drongoski said:
Answer appears to be, taking into account symmetry about the y-axis:

$\100dpi Area\ =\ 2\times \int_{0}^{\frac{\pi }{2}}\ \left [ (\frac{\pi ^{2}}{4}-x^{2})\ -\ (-cos\ x) \right ]\ dx\\ \\ =\ \ 2\times \int_{0}^{\frac{\pi }{2}}\ (\frac{\pi ^{2}}{4}-x^{2}+cos\ x)\ dx\\ \\ =\ \ 2\times \left [ \frac{\pi^{2} }{4}x-\frac{x^{3}}{3}+sin\ x \right ]^{\frac{\pi }{2}}_{0}\ \ =\ \ 2\left ( \frac{\pi ^{2}}{4}\cdot \frac{\pi }{2}-\frac{1}{3}(\frac{\pi }{2})^{3}+sin\frac{\pi }{2} \right )\ -\ 2(0)\\ \\ =\ \ \frac{\pi ^{3}}{6}+2\ \ unit^{2}$

You mean u^2.

. Thankyou Sir. All good.

Areas under/between curves. (1 Viewer)

Psychosomatical

Member

Drongoski

Well-Known Member

Psychosomatical

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)