Calculate the definite integral for 3/x^4 - 1/x^2 + 4 , line y= 2 and y = -2.... Question from @valenciiaa

October 2018 1 15 Report

Calculate the definite integral for 3/x^4 - 1/x^2 + 4 , line y= 2 and y = -2

izharulhaq78 $\displaystyle \int\limits^2_{-2}\frac{3}{x^4}-\frac{1}{x^2}+4\,dx=\int\limits^2_{-2}3x^{-4}-x^{-2}+4\,dx\\\int\limits^2_{-2}\frac{3}{x^4}-\frac{1}{x^2}+4\,dx=\left\frac{3}{-4+1}x^{-4+1}+\frac{-1}{-2+1}x^{-2}+\frac{4}{0+1}x^{0+1}\right|^2_{-2}\\\int\limits^2_{-2}\frac{3}{x^4}-\frac{1}{x^2}+4\,dx=\left\frac{3}{-3}x^{-3}+\frac{-1}{-1}x^{-1}+\frac{4}{1}x^{1}\right|^2_{-2}\\\int\limits^2_{-2}\frac{3}{x^4}-\frac{1}{x^2}+4\,dx=\left-\frac{1}{x^3}+\frac{1}{x}+4x\right|^2_{-2}$

$\displaystyle \int\limits^2_{-2}\frac{3}{x^4}-\frac{1}{x^2}+4\,dx=-\frac{1}{2^3}+\frac{1}{2}+4(2)-\left(-\frac{1}{(-2)^3}+\frac{1}{-2}+4(-2)\right)\\\int\limits^2_{-2}\frac{3}{x^4}-\frac{1}{x^2}+4\,dx=-\frac{1}{8}+\frac{1}{2}+8-\left(\frac{-1}{-8}-\frac{1}{2}-8\right)\\\int\limits^2_{-2}\frac{3}{x^4}-\frac{1}{x^2}+4\,dx=-\frac{1}{8}+\frac{1}{2}+8-\frac{1}{8}+\frac{1}{2}+8\\\int\limits^2_{-2}\frac{3}{x^4}-\frac{1}{x^2}+4\,dx=17-\frac{1}{4}$
$\displaystyle \boxed{\boxed{\int\limits^2_{-2}\frac{3}{x^4}-\frac{1}{x^2}+4\,dx=16\frac{3}{4}}}$

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