What is the integral from 0 to 2 of 2pi(y+1)(1) ?

The integral from 0 to 2 of 2pi(y+1)(1) is 8pi

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integral from 0 to 2 of 2pi(y+1)(1)

\int_{0}^{2} 2 π (y + 1) (1) d y

8 π

Decimal Notation

25.13274 \dots

Solution steps

\int_{0}^{2} 2 π (y + 1) \cdot 1 d y

Take the constant out:

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

= 2 π 1 \cdot \int_{0}^{2} y + 1 d y

Simplify

= 2 π \cdot \int_{0}^{2} y + 1 d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 π (\int_{0}^{2} y d y + \int_{0}^{2} 1 d y)

\int_{0}^{2} y d y = 2

\int_{0}^{2} 1 d y = 2

= 2 π (2 + 2)

Simplify

= 8 π

f (x) = x^{\frac{6}{7}}

f (x) = \frac{3}{x ^{\frac{1}{2}}}, (1, 3)

n = 1 \sum \infty (\frac{4}{5})^{n}

\int_{7}^{8} \frac{7}{50 + x ^{2} - 14 x} d x

\int 2 x^{\frac{5}{2}} d x

What is the integral from 0 to 2 of 2pi(y+1)(1) ?
The integral from 0 to 2 of 2pi(y+1)(1) is 8pi