What is the integral from 1 to 4 of 3/(x^2) ?

The integral from 1 to 4 of 3/(x^2) is 9/4

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integral from 1 to 4 of 3/(x^2)

\int_{1}^{4} \frac{3}{x ^{2}} d x

\frac{9}{4}

Decimal Notation

2.25

Solution steps

\int_{1}^{4} \frac{3}{x ^{2}} d x

Take the constant out:

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

= 3 \cdot \int_{1}^{4} \frac{1}{x ^{2}} d x

Apply the Power Rule

: [- \frac{1}{x}]_{1}^{4}

= 3 [- \frac{1}{x}]_{1}^{4}

Compute the boundaries:

\frac{3}{4}

= 3 \cdot \frac{3}{4}

Simplify

= \frac{9}{4}

\frac{\partial}{\partial x} (x + y + x y + \frac{x}{y} + \frac{y}{x})

y^{^{''}} + y = k

\int_{4}^{5} x^{2} x - 4 d x

\frac{\partial}{\partial x} (\frac{20}{8 + x ^{2} + y ^{2}})

\frac{\partial}{\partial t} (\frac{8.31 t}{V ( t )})

What is the integral from 1 to 4 of 3/(x^2) ?
The integral from 1 to 4 of 3/(x^2) is 9/4