What is the integral from 1 to x of x^2f(t) ?

The integral from 1 to x of x^2f(t) is (x^2f(x^2-1))/2

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integral from 1 to x of x^2f(t)

\int_{1}^{x} x^{2} f (t) d t

\frac{x ^{2} f ( x ^{2} - 1 )}{2}

Solution steps

\int_{1}^{x} x^{2} f t d t

Take the constant out:

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

= x^{2} f \cdot \int_{1}^{x} t d t

Apply the Power Rule

: [\frac{t ^{2}}{2}]_{1}^{x}

= x^{2} f [\frac{t ^{2}}{2}]_{1}^{x}

Compute the boundaries:

\frac{x ^{2}}{2} - \frac{1}{2}

= x^{2} f (\frac{x ^{2}}{2} - \frac{1}{2})

Simplify

= \frac{x ^{2} f ( x ^{2} - 1 )}{2}

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What is the integral from 1 to x of x^2f(t) ?
The integral from 1 to x of x^2f(t) is (x^2f(x^2-1))/2