What is the integral from 3 to x of f(t) ?

The integral from 3 to x of f(t) is (f(x^2-9))/2

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integral from 3 to x of f(t)

\int_{3}^{x} f (t) d t

\frac{f ( x ^{2} - 9 )}{2}

Solution steps

\int_{3}^{x} f t d t

Take the constant out:

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

= f \cdot \int_{3}^{x} t d t

Apply the Power Rule

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

= f [\frac{t ^{2}}{2}]_{3}^{x}

Compute the boundaries:

\frac{x ^{2}}{2} - \frac{9}{2}

= f (\frac{x ^{2}}{2} - \frac{9}{2})

Simplify

= \frac{f ( x ^{2} - 9 )}{2}

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

\int_{0}^{\infty} e^{- 9 y^{2}} d y

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

\int_{0}^{2} x (1 + 2 x^{2})^{3} d x

\int_{0}^{1} x 2^{- x} d x

What is the integral from 3 to x of f(t) ?
The integral from 3 to x of f(t) is (f(x^2-9))/2