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Popular >

integral from 1 to 32 of (sin(\sqrt[5]{x}))/(\sqrt[5]{x^4)}

Solution

Solution

5 (- cos (2) + cos (1))

+1

Decimal Notation

4.78224 \dots

Solution steps

Apply u-substitution

: \int_{1}^{2} 5 sin (u) d u

= \int_{1}^{2} 5 sin (u) d u

Take the constant out:

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

= 5 \cdot \int_{1}^{2} sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= 5 [- cos (u)]_{1}^{2}

Compute the boundaries:

- cos (2) + cos (1)

= 5 (- cos (2) + cos (1))

Popular Examples

limit as x approaches 0 of ((1))/(x^2)

x \to 0 lim (\frac{( 1 )}{x ^{2}})

integral of ((x^2-x+12))/(x^3+6x)

\int \frac{( x ^{2} - x + 12 )}{x ^{3} + 6 x} d x

(1+cos(x))(dy)/(dx)=(1+e^{-y})sin(x)

(1 + cos (x)) \frac{d y}{d x} = (1 + e^{- y}) sin (x)

y^{^{'}} + 2 x y = e^{x}

integral of x^2+9

\int x^{2} + 9 d x

Frequently Asked Questions (FAQ)

What is the integral from 1 to 32 of (sin(\sqrt[5]{x}))/(\sqrt[5]{x^4)} ?
The integral from 1 to 32 of (sin(\sqrt[5]{x}))/(\sqrt[5]{x^4)} is 5(-cos(2)+cos(1))