Answer to Question #178141 in Real Analysis for Izza Eeman

Question

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

Answer 1

Given,

Let $h=f(x)$

As h is an even functioni.e. $f(x)=f(-x)$

$\int _{-\beta}^{\beta}h. dx$

Since Above integral is additive in nature,

So

$\int_{\beta}^{\beta}f(x)dx=\int _{-\beta}^{0}f(x) dx+\int _{0}^{\beta} f(x) dx$

Then, for the first integral we use the expansion/contraction of the interval of integration with k=-1 to get

$\int_{-\beta}^0f(x) dx=-\int_{\beta}^0f(-x)dx$

Since f(x) is an even function by assumption, we have $f(-x) = f(x)$ for all $x \in [0,b]$ . Since$-\int_b^0 = \int_0^b$ we then have,

 $-\int_b^0 f(-x) dx = \int_0^b f(x) dx.$

Putting this all together we have-

$\int_{-\beta}^{\beta}f(x)dx=\int _{0}^{\beta}f(x) dx+\int _{0}^{\beta} f(x) dx=2\int_0^{\beta}f(x) dx$

Hence, $\int_{-\beta}^{\beta}h=2\int_0^{\beta}h$