Question

Help me:
`\int\limits^\pi _0 sin({x}) \, dx`

Answer 1

`\int^(\pi)_(0)sin(x)dx = 2`

Explanation:

To solve for the value of
`\int^(\pi)_(0)sin(x)dx`, we need to first find the antiderivative of
`sin(x)`.

Since the derivative of
`cos(x)` is
`-sin(x)`, therefore the derivative of
`-cos(x)` is
`sin(x)`. With this:

`\int^(\pi)_(0)sin(x)dx = -cos(\pi)-(-cos(0))`

`=-cos(\pi)+cos(0)`

`=-(-1)+1`

`=2`

∴
`\int^(\pi)_(0)sin(x)dx = 2`

Hope this helps :)