Examples & Lab Problems on Integration by Substitution.Ziad Z. AdwanLecture 15 & Lab 1531 / 47

Integration by Substitution [Change of Variables Method]This will be the last technique of Integration in this Lecture. We start bymotivating the technique.Ziad Z. AdwanLecture 15 & Lab 1532 / 47

Integration by Substitution [Change of Variables Method]This will be the last technique of Integration in this Lecture. We start bymotivating the technique.Suppose that you are asked to evaluate the integralZ2xpx2+1dx.(±)Ziad Z. AdwanLecture 15 & Lab 1532 / 47

Integration by Substitution [Change of Variables Method]This will be the last technique of Integration in this Lecture. We start bymotivating the technique.Suppose that you are asked to evaluate the integralZ2xpx2+1dx.(±)After some training, you should be able to see that theintegrand2xpx2+1contains both a functionf(x)& its derivativef0(x). What are they?Ziad Z. AdwanLecture 15 & Lab 1532 / 47

Integration by Substitution [Change of Variables Method]This will be the last technique of Integration in this Lecture. We start bymotivating the technique.Suppose that you are asked to evaluate the integralZ2xpx2+1dx.(±)After some training, you should be able to see that theintegrand2xpx2+1contains both a functionf(x)& its derivativef0(x). What are they?Answer:They aref(x) =x2+1&f0(x) =2x.Ziad Z. AdwanLecture 15 & Lab 1532 / 47

Integration by Substitution [Change of Variables Method]This will be the last technique of Integration in this Lecture. We start bymotivating the technique.Suppose that you are asked to evaluate the integralZ2xpx2+1dx.(±)After some training, you should be able to see that theintegrand2xpx2+1contains both a functionf(x)& its derivativef0(x). What are they?Answer:They aref(x) =x2+1&f0(x) =2x.Go to the next page to see a step-by-step magical procedure for evaluatingthe integral in(±).Ziad Z. AdwanLecture 15 & Lab 1532 / 47

Integration by Substitution [Step-by-Step Procedure]Now, we are ready to learn the technique of integration by substitution.1Letu=x2+1(~).Thendudx=2x=)du=2xdx.Ziad Z. AdwanLecture 15 & Lab 1533 / 47

Integration by Substitution [Step-by-Step Procedure]Now, we are ready to learn the technique of integration by substitution.1Letu=x2+1(~).Thendudx=2x=)du=2xdx.2Can we rewrite our original integralZ2xpx2+1dxin terms ofuonly?Ziad Z. AdwanLecture 15 & Lab 1533 / 47

Integration by Substitution [Step-by-Step Procedure]Now, we are ready to learn the technique of integration by substitution.1Letu=x2+1(~).Thendudx=2x=)du=2xdx.2Can we rewrite our original integralZ2xpx2+1dxin terms ofuonly?Answer:Yes, we haveZ2xpx2+1dx=Zpudu(±±).Ziad Z. AdwanLecture 15 & Lab 1533 / 47

Integration by Substitution [Step-by-Step Procedure]Now, we are ready to learn the technique of integration by substitution.