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Section 5-7 : Computing Definite Integrals

15. Evaluate the following integral, if possible. If it is not possible clearly explain why it is not possible to evaluate the integral.

\(\displaystyle \int_{0}^{4}{{f\left( t \right)\,dt}}\) where \(f\left( t \right) = \left\{ {\begin{array}{*{20}{c}}{2t}&{t > 1}\\{1 - 3{t^2}}&{t \le 1}\end{array}} \right.\)

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Hint : Recall that integrals we can always “break up” an integral as follows, \[\int_{a}^{b}{{f\left( x \right)\,dx}} = \int_{a}^{c}{{f\left( x \right)\,dx}} + \int_{c}^{b}{{f\left( x \right)\,dx}}\]

See if you can find a good choice for “\(c\)” that will make this integral doable.

Start Solution

This integral can’t be done as a single integral give the obvious change of the function at \(t = 1\) which is in the interval over which we are integrating. However, recall that we can always break up an integral at any point and \(t = 1\) seems to be a good point to do this.

Breaking up the integral at \(t = 1\) gives,

\[\int_{0}^{4}{{f\left( t \right)\,dt}} = \int_{0}^{1}{{f\left( t \right)\,dt}} + \int_{1}^{4}{{f\left( t \right)\,dt}}\]

So, in the first integral we have \(0 \le t \le 1\) and so we can use \(f\left( t \right) = 1 - 3{t^2}\) in the first integral. Likewise, in the second integral we have \(1 \le t \le 4\) and so we can use \(f\left( t \right) = 2t\) in the second integral.

Making these function substitutions gives,

\[\int_{0}^{4}{{f\left( t \right)\,dt}} = \int_{0}^{1}{{1 - 3{t^2}\,dt}} + \int_{1}^{4}{{2t\,dt}}\] Show Step 2

All we need to do at this point is evaluate each integral. Here is that work.

\[\int_{0}^{4}{{f\left( t \right)\,dt}} = \int_{0}^{1}{{1 - 3{t^2}\,dt}} + \int_{1}^{4}{{2t\,dt}} = \left. {\left( {t - {t^3}} \right)} \right|_0^1 + \left. {{t^2}} \right|_1^4 = \left[ {0 - 0} \right] + \left[ {16 - 1} \right] = \require{bbox} \bbox[2pt,border:1px solid black]{{15}}\]