1 evaluate the integral remember to use absolute values...

A. Evaluate the integral ∫(5 / ((x-1)(x^2+4))) dx B. Evaluate the integral ∫((x^2-x+20) / (x^3+5x)) dx C. Evaluate the integral ∫(13 / (x^3-27)) dx 2. Make a substitution to express the integrand as a rational function and then evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.) Evaluate the integral ∫(e^x / ((e^x-5)(e^2x +1))) dx 3. Find the area of the region under the given curve from 1 to 3. y = (x^2+7) / (4x-x^2)

Transcript

00:01 In this given question, you have to find the equilibrium equation.

00:06 So equilibrium equation can be find it, find when we have a condition as d.

00:22 Y by d x is equal to f of y should be equal to 0.

00:30 And therefore, here as given the differential equation, the given differential equation, equation is y dash is equal to y r minus 16 and why i'm sorry this is y square minus 16 within the bracket y minus 3 to the whole square this is our equation which need to be differentiated so this will be as y square minus 16 y minus 3 square is equal to 0 and therefore i will be for y minus 3 to the whole square will be equal to 0 and similarly we can say y minus 16 is also equal to 0 so then if y minus 3 to the whole power 2 is equal to 0 that means y minus 3 is equal to 0 and here y is equal to 3 and on the other hand for the second factor which is y squared minus 16 is equal to 0 that means y square is equal to 16 and y is equal to plus minus 4 this is a second value for y and therefore equilibrium points are you can say that our first point will be 3 then we have plus 4 and minus 4 so we can write this in the ascending order as minus 4 3 in minus plus 4 so these are the given points and now we can construct our graph so here's the x's coordinate axis this is our x -axis here we have y -axis and this is our origin so these points will be lie on as you know this is our negative x -axis over here positive xxx is above the origin below the origin we have negative y xxas so these are the respective points 1 2 3 4 then 1 2 3 4 here we have minus 1 minus 2 minus 3 and minus 4 on the x xx similarly minus 1 minus 2 minus 3 minus 4 on y xxas and let's locate our points which we have finite so minus 4 is here this is our point which is minus 4 then we have plus 3 so this is overhead plus 3 and plus 4 is here so we can see here as at 4 3 and minus 4 the y dash value is going upwards is going upwards as you can see here our value is moving from minus 4 to 3 to 4 plus 4...

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