Question

Suppose that 𝛽 > 0 and that ℎ ∈ 𝑅[−𝛽, 𝛽]. 1) If ℎ is even, show that t ∫ h(integration limit from [−𝛽, 𝛽]) = 2 ∫ h(integration limit from [0, 𝛽])

Answer 1

We should prove that for even function h

$\int\limits_{-\beta}^{\beta} h(t)\,dt= 2\int \limits_{0}^{\beta} h(t)\,dt.$

We know that h(-x)=h(x).

Let us divide the integral into two parts:

$\int\limits_{-\beta}^{\beta} h(t)\,dt= \int\limits_{-\beta}^{0} h(t)\,dt + \int\limits_{0}^{\beta} h(t)\,dt$ .

We may see that the first integral may be transformed

$\int\limits_{-\beta}^{0} h(t)\,dt = \Big| x = -t \Big| = -\int\limits_{\beta}^{0} h(-x)\,dx = -\int\limits_{\beta}^{0} h(x)\,dx$ because h(-x)=h(x).

And $-\int\limits_{\beta}^{0} h(x)\,dx = \int\limits_{0}^{\beta} h(x)\,dx$ .

Therefore, $\int\limits_{-\beta}^{\beta} h(t)\,dt= \int\limits_{0}^{\beta} h(x)\,dx + \int\limits_{0}^{\beta} h(t)\,dt = 2\int\limits_{0}^{\beta} h(t)\,dt$ .