What is the integral from 0 to pi of 1 (sin(x)) ?

The integral from 0 to pi of 1 (sin(x)) is d*2

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integral from 0 to pi of 1 (sin(x))

\int_{0}^{π} \frac{d}{1} (sin (x)) d x

d \cdot 2

Solution steps

\int_{0}^{π} \frac{d}{1} sin (x) d x

Take the constant out:

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

= d \cdot \int_{0}^{π} sin (x) d x

Use the common integral:

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

= d [- cos (x)]_{0}^{π}

Compute the boundaries:

2

= d \cdot 2

\int_{0}^{4} ln (3 x + 1) d x

\int_{- 4}^{3} ∣ x + 2 ∣ d x

\int_{1}^{5} \frac{6}{x} d x

\int_{0}^{1} 3 x^{2} (1 + x^{3})^{5} d x

\int_{2}^{4} k (1 - (x - 3)^{2}) d x

What is the integral from 0 to pi of 1 (sin(x)) ?
The integral from 0 to pi of 1 (sin(x)) is d*2