What is the integral from 1 to 6 of kx ?

The integral from 1 to 6 of kx is (35k)/2

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integral from 1 to 6 of kx

\int_{1}^{6} k x d x

\frac{35 k}{2}

Solution steps

\int_{1}^{6} k x d x

Take the constant out:

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

= k \cdot \int_{1}^{6} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{1}^{6}

= k [\frac{x ^{2}}{2}]_{1}^{6}

Compute the boundaries:

\frac{35}{2}

= k \frac{35}{2}

Simplify

= \frac{35 k}{2}

x \to - 2 lim (\frac{x ^{2} - 5}{2 - x})

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

\int_{0}^{1} 6 x d x

y = e^{x}, y = x^{2} - 1, x = - 1, x = 1

n = 1 \sum \infty n e^{6 n}

What is the integral from 1 to 6 of kx ?
The integral from 1 to 6 of kx is (35k)/2