Calculus I - Computing Definite Integrals

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Section 5.7 : Computing Definite Integrals

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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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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.

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