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)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Compute the boundaries:

\frac{1023}{10}

= f \frac{1023}{10}

Simplify

= \frac{1023 f}{10}

y = 5 x^{3} - 5 x, (- 1, 0)

\frac{1000}{s + 1000}

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

\int arctan (8 x) d x

\frac{d}{d x} (7^{x} ln (7))

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)