Very easy integration question
When you integrate x with an upper bound of 1 and a lower bound of minus 1.
You get x^2/2 with an upper limit of 1 and a lower limit of -1
therefore you get
(1/2) - (-1/2)
Which surely give you
1/2+1/2 which equals 1.
However the answer is 0. Why is this so??
What am i doing wrong.
You get x^2/2 with an upper limit of 1 and a lower limit of -1
therefore you get
(1/2) - (-1/2)
Which surely give you
1/2+1/2 which equals 1.
However the answer is 0. Why is this so??
What am i doing wrong.
Original post by anonstudent1
When you integrate x with an upper bound of 1 and a lower bound of minus 1.
You get x^2/2 with an upper limit of 1 and a lower limit of -1
therefore you get
(1/2) - (-1/2)
Which surely give you
1/2+1/2 which equals 1.
However the answer is 0. Why is this so??
What am i doing wrong.
You get x^2/2 with an upper limit of 1 and a lower limit of -1
therefore you get
(1/2) - (-1/2)
Which surely give you
1/2+1/2 which equals 1.
However the answer is 0. Why is this so??
What am i doing wrong.
your (1/2) - (-1/2) should be (1/2) - (1/2) because (-1)^2 = 1
Original post by raees
your (1/2) - (-1/2) should be (1/2) - (1/2) because (-1)^2 = 1
Obviously!! Lack of sleep has turned my brain to mush
Y = f(x) = x
Draw the function. i.e. a straight line y=mx+c where the gradient m is 1 and c is 0.
The integral between the limits is therefore two RA triangles each of of area (base x height)/2.
The definite integral between the limits -1 and +1 is both above and below the x-axis so when x is negative you get an area which has a -ve value.
The square of the integral changes the sign convention so you lose the visual interpretation of the correct result.
To get around this problem perform the integral in two summed (added) parts with limits -1 and 0 and again 0 and +1 which produces:
[1/2 - 0] + [0 - 1/2] = 1/2 - 1/2 = 0.
This is a lesson to be learned when performing definite integrals, sketch the function between the limits to visualise what's happening.
Draw the function. i.e. a straight line y=mx+c where the gradient m is 1 and c is 0.
The integral between the limits is therefore two RA triangles each of of area (base x height)/2.
The definite integral between the limits -1 and +1 is both above and below the x-axis so when x is negative you get an area which has a -ve value.
The square of the integral changes the sign convention so you lose the visual interpretation of the correct result.
To get around this problem perform the integral in two summed (added) parts with limits -1 and 0 and again 0 and +1 which produces:
[1/2 - 0] + [0 - 1/2] = 1/2 - 1/2 = 0.
This is a lesson to be learned when performing definite integrals, sketch the function between the limits to visualise what's happening.
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