What is the integral from 2 to infinity of 9/(2x^2) ?

The integral from 2 to infinity of 9/(2x^2) is 9/4

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integral from 2 to infinity of 9/(2x^2)

\int_{2}^{\infty} \frac{9}{2 x ^{2}} d x

\frac{9}{4}

Decimal Notation

2.25

Solution steps

\int_{2}^{\infty} \frac{9}{2 x ^{2}} d x

Compute the indefinite integral:

\int \frac{9}{2 x ^{2}} d x = - \frac{9}{2 x} + C

Compute the boundaries:

\int_{2}^{\infty} \frac{9}{2 x ^{2}} d x = 0 - (- \frac{9}{4})

= 0 - (- \frac{9}{4})

Simplify

= \frac{9}{4}

\int 5 x cos (x) d x

\frac{( s + 2 )}{( s ^{2} + 3 s + 1 )}

\int \frac{5 x ^{3}}{( 5 - 3 x ^{4} ) ^{8}} d x

\frac{\partial}{\partial x} (sin (x^{2} + y^{2}))

x \to 2 + lim (3 + ∣ 2 x - 4 ∣)

What is the integral from 2 to infinity of 9/(2x^2) ?
The integral from 2 to infinity of 9/(2x^2) is 9/4