What is the integral from 0 to x of tf(x^2-t^2) ?

The integral from 0 to x of tf(x^2-t^2) is f((x^4)/2-(x^4)/4)

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integral from 0 to x of tf(x^2-t^2)

\int_{0}^{x} t f (x^{2} - t^{2}) d t

f (\frac{x ^{4}}{2} - \frac{x ^{4}}{4})

Solution steps

\int_{0}^{x} t f (x^{2} - t^{2}) d t

Take the constant out:

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

= f \cdot \int_{0}^{x} (x^{2} - t^{2}) t d t

Expand

(x^{2} - t^{2}) t : x^{2} t - t^{3}

= f \cdot \int_{0}^{x} x^{2} t - t^{3} d t

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_{0}^{x} x^{2} t d t - \int_{0}^{x} t^{3} d t)

\int_{0}^{x} x^{2} t d t = \frac{x ^{4}}{2}

\int_{0}^{x} t^{3} d t = \frac{x ^{4}}{4}

= f (\frac{x ^{4}}{2} - \frac{x ^{4}}{4})

What is the integral from 0 to x of tf(x^2-t^2) ?
The integral from 0 to x of tf(x^2-t^2) is f((x^4)/2-(x^4)/4)