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

The integral from 1 to x^2 of f(t) is (f(x^4-1))/2

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

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

\frac{f ( x ^{4} - 1 )}{2}

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{x ^{4}}{2} - \frac{1}{2}

= f (\frac{x ^{4}}{2} - \frac{1}{2})

Simplify

= \frac{f ( x ^{4} - 1 )}{2}

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What is the integral from 1 to x^2 of f(t) ?
The integral from 1 to x^2 of f(t) is (f(x^4-1))/2