What is the integral from x to 1 of 6x ?

The integral from x to 1 of 6x is 3(-x^2+1)

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integral from x to 1 of 6x

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

3 (- x^{2} + 1)

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{1}{2} - \frac{x ^{2}}{2}

= 6 (\frac{1}{2} - \frac{x ^{2}}{2})

Simplify

= 3 (- x^{2} + 1)

\int_{0}^{1} \frac{1}{x ^{q}} d x

\int_{0}^{2} 1 + (- 2 x)^{2} d x

\int_{0}^{2} (e^{x} - 1) d x

\int_{7}^{8} 6 lo g_{9} (t) d t

What is the integral from x to 1 of 6x ?
The integral from x to 1 of 6x is 3(-x^2+1)