What is the integral from 1 to 4 of f(x+6) ?

The integral from 1 to 4 of f(x+6) is 51/2 f

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integral from 1 to 4 of f(x+6)

\int_{1}^{4} f (x + 6) d x

\frac{51}{2} f

Solution steps

\int_{1}^{4} f (x + 6) d x

Take the constant out:

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

= f \cdot \int_{1}^{4} 6 + x 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

= f (\int_{1}^{4} 6 d x + \int_{1}^{4} x d x)

\int_{1}^{4} 6 d x = 18

\int_{1}^{4} x d x = \frac{15}{2}

= f (18 + \frac{15}{2})

Join

18 + \frac{15}{2} : \frac{51}{2}

= \frac{51}{2} f

\int_{- 1}^{0} \frac{x ^{2} + 4 x - 1}{x + 2} d x

\int_{1}^{2} \frac{ln ( 2 x )}{x} d x

\int_{0}^{2} ln (x + 4) d x

\int_{0}^{1} x 3 x + 1 d x

\int_{0}^{l n (a)} 6 e^{2 x} d x

What is the integral from 1 to 4 of f(x+6) ?
The integral from 1 to 4 of f(x+6) is 51/2 f