What is the integral from 1 to 2 of Inx ?

The integral from 1 to 2 of Inx is (3In)/2

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integral from 1 to 2 of Inx

\int_{1}^{2} I n x d x

\frac{3 I n}{2}

Solution steps

\int_{1}^{2} I n x d x

Take the constant out:

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

= I n \cdot \int_{1}^{2} x d x

Apply the Power Rule

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

= I n [\frac{x ^{2}}{2}]_{1}^{2}

Compute the boundaries:

\frac{3}{2}

= I n \frac{3}{2}

Simplify

= \frac{3 I n}{2}

y^{^{''}} + 2 y^{^{'}} + 6 y = 0

y^{^{''}} + 25 y = 20 sin (5 t)

2 s

(2 x + y) d x - x d y = 0

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

What is the integral from 1 to 2 of Inx ?
The integral from 1 to 2 of Inx is (3In)/2