What is the integral from 1 to 9 of 1/(2x^2) ?

The integral from 1 to 9 of 1/(2x^2) is 4/9

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integral from 1 to 9 of 1/(2x^2)

\int_{1}^{9} \frac{1}{2 x ^{2}} d x

\frac{4}{9}

Decimal Notation

0.44444 \dots

Solution steps

\int_{1}^{9} \frac{1}{2 x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{1}^{9} \frac{1}{x ^{2}} d x

Apply the Power Rule

: [- \frac{1}{x}]_{1}^{9}

= \frac{1}{2} [- \frac{1}{x}]_{1}^{9}

Compute the boundaries:

\frac{8}{9}

= \frac{1}{2} \cdot \frac{8}{9}

Simplify

= \frac{4}{9}

x^{2} y^{^{''}} + 2 x y^{^{'}} - 12 y = 0, y (1) = - 1, y^{^{'}} (1) = 10

\frac{d x}{d y} = - \frac{4 y ^{2} + 6 x y}{3 y ^{2} + 2 x}

x \to 2 lim (x^{4})

f (x) = \frac{2}{x - 3}

y = 1 - x^{2}, at x = \frac{1}{2}

What is the integral from 1 to 9 of 1/(2x^2) ?
The integral from 1 to 9 of 1/(2x^2) is 4/9