What is the integral from 0 to 2sin(θ) of r ?

The integral from 0 to 2sin(θ) of r is 2sin^2(θ)

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integral from 0 to 2sin(θ) of r

\int_{0}^{2 s i n (θ)} r d r

2 sin^{2} (θ)

Solution steps

\int_{0}^{2 s i n (θ)} r d r

Apply the Power Rule

: [\frac{r ^{2}}{2}]_{0}^{2 s i n (θ)}

= [\frac{r ^{2}}{2}]_{0}^{2 s i n (θ)}

Compute the boundaries:

2 sin^{2} (θ)

= 2 sin^{2} (θ)

\int_{- 3}^{- 2} (2 x - 1) d x

\int_{1}^{5} (- x + 10) d x

\int_{- 1}^{0} ∣ 4 x + 3 ∣ d x

\int_{1}^{2.8} ln (x^{2}) d x

\int_{- 1}^{2} 3 x - 3 d x

What is the integral from 0 to 2sin(θ) of r ?
The integral from 0 to 2sin(θ) of r is 2sin^2(θ)