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First see if the integrand ie x2-4 is negative anywhere in the range (it might change sign either side of anyplace where the function is zero, so first solve for x2-4=0).
If it is, reverse the sign in that part. Here if x<2 it is negative so the modulus gives 4-x2. So integrate 4-x2 from 0 to 2 and then integrate x2-4 from 2 to 3. Add the 2 results together.
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Why derivative of mod x does not exist graphically?
Because mod(x) is not "smooth at x = 0.Suppose f(x) = mod(x). Then f'(x), if it existed, would be the limit, as dx tends to 0, of [f(x+dx) - f(x)]/dx= limit, as dx tends to o , of [mod(x+dx) - mod(x)]/dxWhen x = 0, this simplifies to mod(dx)/dxIf dx > 0 then f'(x) = -1andif dx < 0 then f'(x) = +1Consequently f'(0) does not exist and hence the derivative of mod(x) does not exist at x = 0.Graphically, it is because at x = 0 the graph is not smooth but has an angle.Because mod(x) is not "smooth at x = 0.Suppose f(x) = mod(x). Then f'(x), if it existed, would be the limit, as dx tends to 0, of [f(x+dx) - f(x)]/dx= limit, as dx tends to o , of [mod(x+dx) - mod(x)]/dxWhen x = 0, this simplifies to mod(dx)/dxIf dx > 0 then f'(x) = -1andif dx < 0 then f'(x) = +1Consequently f'(0) does not exist and hence the derivative of mod(x) does not exist at x = 0.Graphically, it is because at x = 0 the graph is not smooth but has an angle.Because mod(x) is not "smooth at x = 0.Suppose f(x) = mod(x). Then f'(x), if it existed, would be the limit, as dx tends to 0, of [f(x+dx) - f(x)]/dx= limit, as dx tends to o , of [mod(x+dx) - mod(x)]/dxWhen x = 0, this simplifies to mod(dx)/dxIf dx > 0 then f'(x) = -1andif dx < 0 then f'(x) = +1Consequently f'(0) does not exist and hence the derivative of mod(x) does not exist at x = 0.Graphically, it is because at x = 0 the graph is not smooth but has an angle.Because mod(x) is not "smooth at x = 0.Suppose f(x) = mod(x). Then f'(x), if it existed, would be the limit, as dx tends to 0, of [f(x+dx) - f(x)]/dx= limit, as dx tends to o , of [mod(x+dx) - mod(x)]/dxWhen x = 0, this simplifies to mod(dx)/dxIf dx > 0 then f'(x) = -1andif dx < 0 then f'(x) = +1Consequently f'(0) does not exist and hence the derivative of mod(x) does not exist at x = 0.Graphically, it is because at x = 0 the graph is not smooth but has an angle.
Find the exact value of the area under the first arch of fx x sinx The first arch is between x 0 and the first positive x-intercept?
First, find the upper limit of integration by setting xsin(x)=0. It should be pi. Then use integration by parts to integrate xsin(x) from 0 to pi u=x dv=sinx dx du=dx v=-cosx evaluate the -xcosx+sinx from 0 to pi the answer is pi ps webassign sucks
Find the derivative for y equals 5-2x?
7
What is the integral of 2 times 1-x dx with limits 0 to t?
,/` 2(1 - x) dx,/` 2 - 2x dx2x - x2 ...evaluated from 0 to t gives us...2t - t2 - [2(0) - (0)2]2t - t2
What is a derivative of inspired?
Inspired could be though of as an integer as it does not have an exponent. d/dx(inspired) = 0 ==== Or, as a variable with the implied exponent 1. Using the power rule. d/dx(inspired 1 - 1) = inspired0 = 1 ====
