Integration Q's: King's Property (1 Viewer)

hmim · Apr 30, 2023

Could Someone please show me how to do these integration questions with King's Property, please?

gazzaboy · May 1, 2023

I hadn't heard of King's Property but a quick Google says that it is
$\int_a^b f(x) \mathrm{d} x = \int_a^b f(a+b-x) \mathrm{d}x$
You prove this using integration by substitution. The version that you use for all of these questions is:
$\int_{-a}^a f(x) \mathrm{d} x = \int_{-a}^a f(-x) \mathrm{d} x.$

For the first one, let I equal to the value of the definite integral. By King's Property, you can say that
$\int_{-1}^1 \frac{x^2}{1+e^x} \mathrm{d} x = \int_{-1}^1 \frac{x^2}{1+e^{-x}} \mathrm{d} x = \int_{-1}^1 \frac{x^2 e^x}{1+e^{x}} \mathrm{d} x.$
You can then add two of I together to get:
$2I = \int_{-1}^1 \frac{x^2}{1+e^x} \mathrm{d} x + \int_{-1}^1 \frac{x^2 e^x}{1+e^{x}} \mathrm{d} x = \int_{-1}^1 x^2 \mathrm{d} x = \left[ \frac{x^3}{3} \right]_{-1}^1 = \frac{2}{3} \Rightarrow I = \frac{1}{3}.$

For the second one, you apply King's property to show that
$\int_{-\pi/4}^{\pi/4} \frac{1}{1+\sin(x)} \mathrm{d}x = \int_{-\pi/4}^{\pi/4} \frac{1}{1+\sin(-x)} \mathrm{d}x = \int_{-\pi/4}^{\pi/4} \frac{1}{1-\sin(x)} \mathrm{d}x.$
If J is the value of the definite integral, then you can say that
$2J = \int_{-\pi/4}^{\pi/4} \frac{1}{1+\sin(x)} \mathrm{d}x + \int_{-\pi/4}^{\pi/4} \frac{1}{1-\sin(x)} \mathrm{d}x = \int_{-\pi/4}^{\pi/4} \frac{2}{1-\sin^2(x)} \mathrm{d}x = \int_{-\pi/4}^{\pi/4} 2 \sec^2 (x) \mathrm{d}x = 2 \left[ \tan(x)\right]_{-\pi/4}^{\pi/4} = 4 \Rightarrow J = 2.$

For the last one, it's very similar so I'll skip some steps. If K is the value of the definite integral, then
$2K = \int_{-\pi/2}^{\pi/2} \sin^2 (x) \mathrm{d} x = \frac{\pi}{2} \Rightarrow K = \frac{\pi}{4}.$

hmim · May 1, 2023

gazzaboy said:
I hadn't heard of King's Property but a quick Google says that it is
$\int_a^b f(x) \mathrm{d} x = \int_a^b f(a+b-x) \mathrm{d}x$
You prove this using integration by substitution. The version that you use for all of these questions is:
$\int_{-a}^a f(x) \mathrm{d} x = \int_{-a}^a f(-x) \mathrm{d} x.$

For the first one, let I equal to the value of the definite integral. By King's Property, you can say that
$\int_{-1}^1 \frac{x^2}{1+e^x} \mathrm{d} x = \int_{-1}^1 \frac{x^2}{1+e^{-x}} \mathrm{d} x = \int_{-1}^1 \frac{x^2 e^x}{1+e^{x}} \mathrm{d} x.$
You can then add two of I together to get:
$2I = \int_{-1}^1 \frac{x^2}{1+e^x} \mathrm{d} x + \int_{-1}^1 \frac{x^2 e^x}{1+e^{x}} \mathrm{d} x = \int_{-1}^1 x^2 \mathrm{d} x = \left[ \frac{x^3}{3} \right]_{-1}^1 = \frac{2}{3} \Rightarrow I = \frac{1}{3}.$

For the second one, you apply King's property to show that
$\int_{-\pi/4}^{\pi/4} \frac{1}{1+\sin(x)} \mathrm{d}x = \int_{-\pi/4}^{\pi/4} \frac{1}{1+\sin(-x)} \mathrm{d}x = \int_{-\pi/4}^{\pi/4} \frac{1}{1-\sin(x)} \mathrm{d}x.$
If J is the value of the definite integral, then you can say that
$2J = \int_{-\pi/4}^{\pi/4} \frac{1}{1+\sin(x)} \mathrm{d}x + \int_{-\pi/4}^{\pi/4} \frac{1}{1-\sin(x)} \mathrm{d}x = \int_{-\pi/4}^{\pi/4} \frac{2}{1-\sin^2(x)} \mathrm{d}x = \int_{-\pi/4}^{\pi/4} 2 \sec^2 (x) \mathrm{d}x = 2 \left[ \tan(x)\right]_{-\pi/4}^{\pi/4} = 4 \Rightarrow J = 2.$

For the last one, it's very similar so I'll skip some steps. If K is the value of the definite integral, then
$2K = \int_{-\pi/2}^{\pi/2} \sin^2 (x) \mathrm{d} x = \frac{\pi}{2} \Rightarrow K = \frac{\pi}{4}.$

Thank you so much for your help!!

Make a Difference – Donate to Bored of Studies!

Integration Q's: King's Property (1 Viewer)

hmim

Member

gazzaboy

Member

hmim

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)