What is the integral from 110 to 160 of (160)/(3x^2) ?

The integral from 110 to 160 of (160)/(3x^2) is 5/33

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integral from 110 to 160 of (160)/(3x^2)

\int_{110}^{160} \frac{160}{3 x ^{2}} d x

\frac{5}{33}

Decimal Notation

0.15151 \dots

Solution steps

\int_{110}^{160} \frac{160}{3 x ^{2}} d x

Take the constant out:

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

= \frac{160}{3} \cdot \int_{110}^{160} \frac{1}{x ^{2}} d x

Apply the Power Rule

: [- \frac{1}{x}]_{110}^{160}

= \frac{160}{3} [- \frac{1}{x}]_{110}^{160}

Compute the boundaries:

\frac{1}{352}

= \frac{160}{3} \cdot \frac{1}{352}

Simplify

= \frac{5}{33}

\frac{d}{d x} (\frac{3}{x - 2})

y = cos (a^{3} + x^{3})

\frac{d y}{d x} = - 5 e^{x + y}

\int_{1}^{4} (3 x^{2} + x) d x

\frac{\partial}{\partial x} (cos (x z) + sin (yz) - z)

What is the integral from 110 to 160 of (160)/(3x^2) ?
The integral from 110 to 160 of (160)/(3x^2) is 5/33