Why does $\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx$

Question

Yeah, so why does $ \displaystyle \int_a^b f(x) \ dx = \displaystyle \int_a^b f(a+b-x) \ dx$?

Also, where can I read up about exploiting the symmetry of trig functions whilst integrating?

Answer 1

Using the substitution $t = a + b - x, \mathrm dt = -\mathrm dx$, $$ \int_a^b f(x)\;\mathrm dx = -\int_b^a f(t)\;\mathrm dt = \int_a^b f(t)\;\mathrm dt $$

Answer 2

$u=a+b-x$, $x=a$ implies $u=b, x=b$ implies, $u=a$, $du=-dx$, $\int_a^bf(x)dx=\int_b^af(a+b-u)(-du)=\int_a^bf(a+b-u)du$.

Answer 3

If $f(x)$ is continuous, then by the FTOC, we have

$$\int_a^bf(a+b-x)\ dx=-F(a+b-b)-(-F(a+b-a))=F(b)-F(a)=\int_a^bf(x)\ dx$$

Answer 4

Let $u=a+b-x$. At $x=a$, we get $u=b$. At $x=b$, we get $u=a$. Also, $du=-dx$. Thus $$\int_{a}^bf(a+b-x)dx=\int_{b}^af(u)(-du)=-\int_{b}^af(u)du=\int_{a}^bf(u)du.$$ Since we are dealing with definite integrals,we have $$\int_{a}^bf(u)du=\int_{a}^bf(x)dx.$$ Done. Hope it helps.