What is the integral from 1 to 2 of t^4f(t^5) ?

The integral from 1 to 2 of t^4f(t^5) is (1023f)/(10)

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integral from 1 to 2 of t^4f(t^5)

\int_{1}^{2} t^{4} f (t^{5}) d t

\frac{1023 f}{10}

Solution steps

\int_{1}^{2} t^{4} f t^{5} d t

Take the constant out:

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

= f \cdot \int_{1}^{2} t^{4} t^{5} d t

Simplify

t^{4} t^{5} : t^{9}

= f \cdot \int_{1}^{2} t^{9} d t

Apply the Power Rule

= f [\frac{t ^{10}}{10}]_{1}^{2}

Compute the boundaries:

\frac{1023}{10}

= f \frac{1023}{10}

Simplify

= \frac{1023 f}{10}

t an g e n t y = 5 x^{3} - 5 x, (- 1, 0)

in v erse l a pl a ce \frac{1000}{s + 1000}

\frac{d}{d x} (\frac{2 2}{x ^{2}})

d er i v a t i v e \frac{3 e ^{x}}{( 3 + e ^{x} ) ^{2}}

\frac{\partial}{\partial y} (5 + x^{2} + y^{2} + 2 x y)

What is the integral from 1 to 2 of t^4f(t^5) ?
The integral from 1 to 2 of t^4f(t^5) is (1023f)/(10)