Area of a Parallelogram with out cross products

c-murda
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Homework Statement



given the vertices:
(0,0)
(3,1)
(2,3)
(5,4)

Homework Equations



solving without cross products and with mathematica if available


The Attempt at a Solution



Te1=[3]
[1]
Te2=[2]
[3]


[det[Ay]]
________

[det[A]]

get you the solution

i got 7

correct?
 
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Yeah it looks ok. By the way since you have mathematica, couldn't you have verified the answer?
 
Defennder said:
Yeah it looks ok. By the way since you have mathematica, couldn't you have verified the answer?

i don't have mathematica that why i was checking my answer. and to know i could do it by hand.

thanks!
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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