What is the integral from 0 to 2 of 3x ?

The integral from 0 to 2 of 3x is 6

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integral from 0 to 2 of 3x

\int_{0}^{2} 3 x d x

6

Solution steps

\int_{0}^{2} 3 x d x

Take the constant out:

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

= 3 \cdot \int_{0}^{2} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{2}

= 3 [\frac{x ^{2}}{2}]_{0}^{2}

Compute the boundaries:

2

= 3 \cdot 2

Simplify

= 6

\int (3 u + \frac{1}{u}) d u

f (x) = (x - 1)^{2} + 1, at x = 1

y^{^{'}} + \frac{1}{x} y = 4 x^{2}

\frac{\partial ^{2}}{\partial x \partial y} (y^{5} sin (3 x))

n = 1 \sum \infty \frac{n 7 ^{n}}{n !}