Just to go into more detail than I did in our PMs and the comments on your last question...
You have to keep in mind that the limits of integration, the interval
, only apply to the original variable of integration (y).
When you make the substitution
, you not only change the variable but also its domain. To find out what the new domain is is a matter of plugging in every value in the y-interval into the substitution relation to find the new t-interval domain for the new variable (t).
After replacing
and the differential
with the new variable
and differential
, you saw that you could reduce the integral to -1. This is a continuous function, so the new domain can be constructed just by considering the endpoints of the y-interval and transforming them into the t-domain.
When
, you have
.
When
, you have
.
Geometrically, this substitution allows you to transform the area as in the image below. Naturally it's a lot easier to find the area under the curve in the second graph than it is in the first.
6 is the square root of 36.
6(6) = 36
Equilateral: "equal"-lateral (lateral means side) so they have all equal sides. Isosceles: means "equal legs", and we have two legs, right? Also iSOSceles has two equal "Sides" joined by an "Odd" side. Scalene: means "uneven" or "odd", so no equal sides.
56.07 < 56.7
Because 56.7 is 56.70