What is the integral from a to b of 1/(x^3) ?

The integral from a to b of 1/(x^3) is -1/(2b^2)+1/(2a^2)

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integral from a to b of 1/(x^3)

\int_{a}^{b} \frac{1}{x ^{3}} d x

- \frac{1}{2 b ^{2}} + \frac{1}{2 a ^{2}}

Solution steps

\int_{a}^{b} \frac{1}{x ^{3}} d x

Apply the Power Rule

: [- \frac{1}{2 x ^{2}}]_{a}^{b}

= [- \frac{1}{2 x ^{2}}]_{a}^{b}

Compute the boundaries:

- \frac{1}{2 b ^{2}} + \frac{1}{2 a ^{2}}

= - \frac{1}{2 b ^{2}} + \frac{1}{2 a ^{2}}

f (x) 50 - x

f (x) = 6 + \frac{2}{x} + \frac{3}{x ^{2}}, x = x

f (x) = 2 x^{2} - 3 x + 3

\int \frac{x ^{2} + 2 x + 3}{x ^{3} + 3 x ^{2} + 9 x} d x

y = x^{2} - 2 x + 3, [- 2, 2]

What is the integral from a to b of 1/(x^3) ?
The integral from a to b of 1/(x^3) is -1/(2b^2)+1/(2a^2)