What is the integral of 3x(7-x) ?

The integral of 3x(7-x) is (21x^2)/2-x^3+C

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integral of 3x(7-x)

\int 3 x (7 - x) d x

\frac{21 x ^{2}}{2} - x^{3} + C

Solution steps

\int 3 x (7 - x) d x

Take the constant out:

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

= 3 \cdot \int x (7 - x) d x

Expand

x (7 - x) : 7 x - x^{2}

= 3 \cdot \int 7 x - x^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int 7 x d x - \int x^{2} d x)

\int 7 x d x = \frac{7 x ^{2}}{2}

\int x^{2} d x = \frac{x ^{3}}{3}

= 3 (\frac{7 x ^{2}}{2} - \frac{x ^{3}}{3})

Simplify

3 (\frac{7 x ^{2}}{2} - \frac{x ^{3}}{3}) : \frac{21 x ^{2}}{2} - x^{3}

= \frac{21 x ^{2}}{2} - x^{3}

Add a constant to the solution

= \frac{21 x ^{2}}{2} - x^{3} + C

f (x) = 5 + 4 x^{2} - 2 x^{3}, (1, 7)

\int \frac{( 3 x ^{2} - 4 ) ^{\frac{3}{2}} 4 x ^{3}}{x ^{2}} d x

\frac{d}{d x} (\frac{sin ( x )}{1 + cos ( x )})

\int e^{- s t} t cos (t) d t

\frac{5}{s ( s ^{2} + s + 1 )}