Question

find value of the integral f [3,1] 1/x^2 dx

Answer 1

\[\int\limits_{1}^{3} \frac{ 1 }{ x^2 } dx\]

\[Deltax=\frac{ 3-1 }{ n }=\frac{ 2 }{ n }\]

Answer 2

\[\int\limits^3_1\frac1{x^2}\mathrm dx\\
=\int\limits^3_1x^{-2}\mathrm dx\\
=\left.\frac{x^{-2+1}}{-2+1}\right|_1^3\\=\]

Answer 3

so the value of the integral is 2/3

as f(3)=-1/3
f(1)=-1
so -(-2/3) which = 2/3

Answer 4

I think you've done this right , because you have the right answer,

but im going to show you the steps i would use anyway

\[\int\limits^3_1\frac1{x^2}\mathrm dx\\
=\int\limits^3_1x^{-2}\mathrm dx\\
=\left.\frac{x^{-2+1}}{-2+1}\right|_1^3\\
=\left.-{x^{-1}}\right|_1^3\\
=-3^{-1}-(-1^{-1})\\=-\frac13+1\\=1-\frac13\\=\frac23\]