explanation with integrals

Question

(a) Let f(x),g(y) be continuous functions defined for a ≤ x ≤ b, c ≤ y ≤ d, resepctively.

Let h(x,y) = f(x)g(y). Let R = {(x,y) : a ≤ x ≤ b,c ≤ y ≤ d}.

Show that ∫_R h(x,y)dA =(∫_a^b f(x)dx)(∫_c^d g(y)dy).

 

(b) For R > 0 let Q_R = {(x,y) : −R ≤ x ≤ R,−R ≤ y ≤ R} (square of sides of length 2R, center at (0,0)).

Show that lim _R→∞∫Q_R e^(−x²−y²) dA =(∫_-∞^∞ e^(-x²)dx)².

(That∫_-∞^∞ e^(-x²)dx exists follows by comparison, for example with∫_-∞^∞ e^(-|x|)dx.)

(c) Let R > 0. Show that if D_R = {(x,y) : x² + y² ≤ R²} is the disc of radius R centered at the origin,

then ∫_QR/√2 e^(−x²−y²) dA ≤∫_DR e^(−x²−y²) dA ≤∫_Q2R e^(−x²−y²) dA

and explain how this, together with the squeeze theorem, shows that one also has

lim _R→∞∫_DR e^(−x²−y²) dA =(∫_-∞^∞ e^(-x²)dx)².

 

(d) Changing variables to polar coordinates compute∫_DR e^(−x²−y²) dA,

use this to compute lim_R→∞∫_DR e^(−x²−y²) dA and establish that ∫_-∞^∞ e^(−x²) dx = √π.

 

_QR is suppose to be Q_R

_QR/√2 is suppose to be Q_R/√2

_DR is suppose to be D_R

 

Really at a loss

Thank you

Answer 1

Anita, I answered this question partialy earlier. Have a look at 

https://www.wyzant.com/resources/answers/461879/show_that_integrals

and if some question, feel free to ask