∫[0,22] x+13 dx
= 1/2 x^2 + 13x [0,22]
= (1/2 * 22^2 + 13*22)-(1/2 * 0^2 + 13*0)
= 528
Evaluate the definite integral.
function: x+13 with respect to variable x
lower limit:0
upper limit:22
4 answers
I tried that way, my answer was wrong
Original problem was: absolute value x-13, because of absolute value I made it x+13
Hello,? By calculus time you should be well aware that |x-13| ≠ x+13
|x-13| makes a big difference.
For x<13, |x-13| = -(x-13)
For x>=13, |x-13| = x-13
So, that means that we have to break the interval into two parts:
∫[0,22] x-13 dx
= ∫[0,13] -(x-13) dx + ∫[13,22] x-13 dx
= ∫[0,13] -x+13 dx + ∫[13,22] x-13 dx
= 125
Check: the area is the sum of two triangles.
One of base 13 and height 13
One of base 9 and height 9
1/2 * 169 + 1/2 * 81 = 125
|x-13| makes a big difference.
For x<13, |x-13| = -(x-13)
For x>=13, |x-13| = x-13
So, that means that we have to break the interval into two parts:
∫[0,22] x-13 dx
= ∫[0,13] -(x-13) dx + ∫[13,22] x-13 dx
= ∫[0,13] -x+13 dx + ∫[13,22] x-13 dx
= 125
Check: the area is the sum of two triangles.
One of base 13 and height 13
One of base 9 and height 9
1/2 * 169 + 1/2 * 81 = 125
2 answers