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Here is the solution If anyone can explain to me, that'll be great!

Also, is THIS an inner space? For me, it comes out not to be. At the step (u+v,w), it gets disproved.

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    You need to format this using \LaTeX. As posed, I can't make heads or tails of what you are asking. The two links do not seem related. – 2017-01-24
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    And post your disproof. – 2017-01-24

1 Answers 1

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For your first question, integration by parts is your friend :

$$\int_0^\pi (x+1) \cos(x) dx = \int_0^\pi x \cos(x) dx + \underbrace{\int_0^\pi \cos(x) dx}_{=0} $$

$$= [ - x \sin(x) ]_0^\pi - \int_0^\pi -\sin(x) dx = 0 - [-\cos(x)]_0^\pi = 1-(-1) = 2 $$

For your second question, it's unrelated to the first, but it's clearly an inner product space (forget about the matrix writting, it's the same as the usual $\Bbb{R}^4$ space)