Calculus II Class 01/27

Calculus I Quiz 0127

Given integrand x^4 e^(2x) we can express the integral in terms of u and dv in the ways indicated below.  Some of these breakdowns, when substituted into the formula, will give us a simpler form on the right-hand side than on the left, while others will result in a more complicated right-hand form.

Integration by parts consists of recognizing and implementing breakdowns which lead to simpler, and ultimately to easily integrable, forms. 

Give all possible substitutions and all possible breakdowns into u and dv, then see which, if any, lead to a closed-form result for each of the following integrals:

• integral ( t^2 e^t dt)

The expression t^2 e^t dt can be expressed in the form u dv in the following ways:

• u = t^2 e^t, dv = dt

• u = t^2, dv = e^t dt

• u = e^t, dv = t^2 dt

• u = t, dv = t e^t dt

• u = t e^t, dv = t dt

• u = 1, dv = t^2 e^t dt.

We can check to see if the integration by parts formula works for each of these breakdowns, as is done for some of the breakdowns below.

The breakdown u = t^2 e^t is seen to result in integral(v du) which has a higher power of t than the original.  We could repeat the process, obtaining higher and higher powers of t, but there is no end to that process and we never get an easily integrable expression.

The breakdown shown below, where we let u = t^2, leads to du = 2 t dt and therefore to an integral with a lower power of t.  

Since lower and lower integer powers will eventually lead to the 0 power, which is a constant, we might be on to something here.  

If we repeat the process on the last integral , of t * e^t with respect to t, we would obtain an expression we could integrate.  This result would then be subsituted into the last step in the preceding figure to obtain the integral of the t^2 e^t expression.

To find the integral of t sin(t) with respect to t, start with u = t and dv = sin(t) dt and obtain the result shown below.

• integral ( sin(x) e^(cos(x)) dx)

The following breakdowns are possible:

• u = sin(x), dv = e^(cos(x)) dx

• u = e^(cos(x)), dv = sin(x) dx

• etc.

Turns out that this one is most easily done using just plain old substitution u = cos(x), so that du = - sin(x) dx.  But see if integration by parts helps.

The remaining breakdowns require tricks and due to weather-related sporadic attendance during the last two classes will not be addressed today.

• integral ( t^4 ( 1 - t)^2 dt )

• integral ( t^3 cos(t^2) dt)