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2 years ago

Multiply the first expression by the second:a2 + b2 + c2 - ab - bc - ca, a + b+ c

Mathematics

2 answers:

Stels [109]2 years ago

8 0

Answer:

https://orion.math.iastate.edu/dept/links/formulas/form1.pdf

Step-by-step explanation:

Fiesta28 [93]2 years ago

8 0

Please find attached photograph for your answer

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3 years ago

<img src="https://tex.z-dn.net/?f=%5Cbegin%7Bequation%7D%5Ctext%20%7B%20Question%3A%20If%20%7D%20%5Cint_%7B%5Cfrac%7B-1%7D%7B%5C

Elodia [21]

We want to evaluate

$\displaystyle \int_{-\frac1{\sqrt2}}^{\frac1{\sqrt2}} \sqrt{\left(\frac{x-1}{x+1}\right)^2 + \left(\frac{x+1}{x-1}\right)^2 - 2} \, dx$

First we note that the integrand is even (replacing x with -x doesn't fundamentally alter the function being integrated), so this is equal to

$\displaystyle 2 \int_0^{\frac1{\sqrt2}} \sqrt{\left(\frac{x-1}{x+1}\right)^2 + \left(\frac{x+1}{x-1}\right)^2 - 2} \, dx$

The radicand reduces significantly to

$\displaystyle \left(\frac{x-1}{x+1}\right)^2 + \left(\frac{x+1}{x-1}\right)^2 - 2 = \frac{16x^2}{(1-x^2)^2}$

so that taking the square root, we simplify the integral to

$\displaystyle 8 \int_0^{\frac1{\sqrt2}} \frac x{1-x^2} \, dx$

which is trivially computed with a substitution of $y = 1 - x^2$ and $dy=-2x\,dx$ :

$\displaystyle -4 \int_1^{\frac12} \frac{dy}y = 4 \int_{\frac12}^1 \frac{dy}y = -4 \ln\left(\frac12\right) = \ln(2^4) = \boxed{\ln(16)}$

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2 years ago

Multiply the first expression by the second:a2 + b2 + c2 - ab - bc - ca, a + b+ c​

Multiply the first expression by the second:a2 + b2 + c2 - ab - bc - ca, a + b+ c