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

The integral from 3x^2-6x to 2x-x^2 of (3x) is 12x^2(-x^2+4x-4)

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integral from 3x^2-6x to 2x-x^2 of (3x)

\int_{3 x^{2} - 6 x}^{2 x - x^{2}} (3 x) d x

12 x^{2} (- x^{2} + 4 x - 4)

Solution steps

\int_{3 x^{2} - 6 x}^{2 x - x^{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_{3 x^{2} - 6 x}^{2 x - x^{2}} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{3 x^{2} - 6 x}^{2 x - x^{2}}

= 3 [\frac{x ^{2}}{2}]_{3 x^{2} - 6 x}^{2 x - x^{2}}

Compute the boundaries:

\frac{( 2 x - x ^{2} ) ^{2}}{2} - \frac{( 3 x ^{2} - 6 x ) ^{2}}{2}

= 3 (\frac{( 2 x - x ^{2} ) ^{2}}{2} - \frac{( 3 x ^{2} - 6 x ) ^{2}}{2})

Simplify

= 12 x^{2} (- x^{2} + 4 x - 4)

\int_{2}^{4} 3 x^{3} d x

\int_{0}^{1} 1 + 4 x d x

\int_{2}^{3} \frac{1}{x + 1} d x

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

\int_{0}^{\frac{π}{6}} 2 cos (3 x) d x

What is the integral from 3x^2-6x to 2x-x^2 of (3x) ?
The integral from 3x^2-6x to 2x-x^2 of (3x) is 12x^2(-x^2+4x-4)