Double integral of e^(x/y) dy dx y=x to 1 x=0 to 1?
Rewrite the region bounded by y = x and y = 1 for x in [0, 1] as x = 0 to x = y for y in [0, 1]. So, the integral equals ∫(y = 0 to 1) ∫(x = 0 to y) e^(x/y) dx dy = ∫(y = 0 to 1) ye^(x/y) {for x = 0 to y} dy = ∫(y = 0 to 1) y (e^1 - e^0) dy = (1/2) y^2 (e - 1) {for y = 0 to 1} = (1/2)(e - 1). I hope this helps!
Using U-Substitution...Definite Integral?
Hey guys :) Can you help me solve this definite integral please? ∫ (x^3+x^4(tanx))dx the upper limit: pi/4 the lower limit: -pi/4 the answer is 0. but I need an explanation, if you can please do so. I started with letting u=x^4 then i got du/dx=4x^3 dx=du/4x^3 but i messed up, I think.. thank you!!