Recall that a circle of radius 2 centered at the origin has equation
where the positive root gives the top half of the circle in the x-y plane. The definite integral corresponds to the area of the right half of this top half. Since the area of a circle with radius
is
, it follows that the area of a quarter-circle would be
.
You have
, so the definite integral is equal to
.
Another way to verify this is to actually compute the integral. Let
, so that
. Now
Recall the half-angle identity for cosine:
This means the integral is equivalent to
Let one integer be x, the consecutive odd integer is x+2
reciprocals are 1/x and 1/(x+2)
1/x +1/(x+2)=24/143
make the denominators the same:
(x+2)/[x(x+2)] +x/[x(x+2)]=24/143
143(2x+2)=24(x)(x+2)
286x+286=24x²+48x
24x²-238x-286=0
use the quadratic formula: x=22 this doesn't work because 22 is not an odd number.
or x=-2.1666666666 (this is not an integer)
weird.
Here's a way to do it.
Let 4e +2 = 5n +1 . . . . . . for some integer n
Then e = (5n -1)/4 = n + (n -1)/4
We want (n-1)/4 to be an integer, so let it be integer m.
... m = (n -1)/4
... 4m = n -1
... 4m +1 = n
Substituting this into our expression for e gives
... e = (5(4m+1) -1)/4 = (20m +4)/4 = 5m +1
e = 5m+1 for any integer m
Answer: x = 6
<u>Step-by-step explanation:</u>
2[3x - (4x - 6)] = 5(x - 6)
2[3x - 4x + 6] = 5x - 30
2[-x + 6] = 5x - 30
-2x + 12 = 5x - 30
<u>+2x </u> <u>+2x </u>
12 = 7x - 30
<u>+30 </u> <u> +30 </u>
42 = 7x
6 = x
Answer:
-6/3
Step-by-step explanation:
Move down 6 from the top point.
Then move right 3 to the bottom point.
