Which integral gives the area of the region in the first quadrant bounded by the axes, y = ex, x = ey, and the line x = 4?

It will be easy if you graph it out.

Note that for x < 1x < 1, log(x) < 0. Therefore, for 0 ≤ x ≤ 1, the area is the area under the curve e^x. For 1 ≤ x ≤, e^x > log(x) ≥.

Putting it together:

∫(x = 0 to 1) e^x dx + ∫(x = 1 to 4) [e^x - ln(x)] dx 
= [e^x] (x = 0 to 1) + [e^x - x(ln(x) - 1)](x = 1 to 4) 
= (e^1 - e^0) + ((e^4 - 4(ln(4) - 1)) - (e^1 - 1(ln(1) - 1))) 
= (e - 1) + (e^4 - 8 ln(2) + 4 - e + 0 - 1) 
= e^4 + 2 - 8 ln(2)

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl