unable to understand tan/arctan-related integral transformation

Question

I'm currently studying area between curve and x axis on Khan Academy.

I understand that the

But I am puzzled by

how did we get to this step from the previous step?

Answer 1

That just says the "area to the left of the symmetry axis that is the $y$-axis is equal to the area to the right of it" which stated mathematically is $\int_{-1}^{0}f(x)\mathrm dx=\int_{0}^{1} f(x)\mathrm dx$ for $f(x)=1/(1+x^2)$ which is why you can rewrite the integral as: $$\int_{-1}^{1}\dfrac{2}{1+x^2}\mathrm dx=\int_{-1}^{0}\dfrac{2}{1+x^2}\mathrm dx+\int_{0}^{1}\dfrac{2}{1+x^2}\mathrm dx=4\int_{0}^{1}\dfrac{1}{1+x^2}\mathrm dx$$ $$\text{ar}(f)_{-1}^{0}=\text{ar}(f)_{0}^{1}$$

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More generally the following holds for some function $g(x)$ provided it is even i.e. $g(-x)=g(x)$: $$\int_{-a}^{a}g(x)\mathrm dx=2\int_{0}^{a}g(x)\mathrm dx$$

Answer 2

We know that, if $f$ is an even function - that is, $f(-x) = f(x)$ for all $x$ - then

$$\int_{-a}^a f(x) dx = 2 \int_0^a f(x)dx$$

The function $$f(x) = \frac{2}{1+x^2}$$ is in fact such a function.

Answer 3

An even function occurs when f(x)=f(-x) and can best be visualized as a function that is symmetric over the x-axis. This means in terms of area, that $\int_0^1 1/(1+x^2) \,dx$ = $\int_{-1}^0 1/(1+x^2) \,dx$. Thus what the problem is saying is that because the function is even, 2$\int_{-1}^1 1/(1+x^2) \,dx$ can more easily be found by finding 2($\int_0^1 1/(1+x^2) \,dx$+$\int_0^1 1/(1+x^2) \,dx$) or 4$\int_0^1 1/(1+x^2) \,dx$.

Answer 4

For an even (integrable) function $f(-x) = f(x)$ you have

• $I = \int_{-a}^0 f(x) dx \stackrel{u=-x}{=} -\int_{a}^0 f(-u) du = \int_{0}^a f(u) du = I$

So, you can split the integral: $\int_{-a}^a f(x) dx = \int_{-a}^0f(x) dx + \int_{0}^a f(x) dx = 2\int_{0}^a f(x) dx$