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The shading in Figure 4.3.1(a) corresponds to the area bounded by the $x$axis and the graph of $y\={x}^{3}$ on the interval $\left[1\,2\right]$. Because part of the shaded region is below the $x$axis, the area is given by

${\int}_{1}^{0}{x}^{3}\mathit{DifferentialD;}xplus;{\int}_{0}^{2}{x}^{2}\mathit{DifferentialD;}x$ = $\frac{{35}}{{12}}$${}$
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Alternatively, evaluate the definite integral:

${\int}_{1}^{2}{x}^{3}\mathit{DifferentialD;}xequals;$ $\frac{{x}^{4}}{4}{\]}_{1}^{2}$ = $\left({2}^{4}{\left(1\right)}^{4}\right)\/4\=15\/4$


Figure 4.3.1(a) Graph of ${x}^{3}$ on $\left[1\,2\right]$






The antiderivative of ${x}^{3}$ is found by the Power rule: add 1 to the power and divide by the new power.
The simplest antiderivative has been used. If the antiderivative were taken as ${x}^{4}\/4\+C$, the result would be the same. The value of $C$ at the upper limit is $C$, as is the value at the lower limit; adding an arbitrary constant to the antiderivative induces $CC\=0$ in the evaluation of the antiderivative at the endpoints.
${\left[\frac{{x}^{4}}{4}\+C\right]}_{1}^{2}\=\left(\frac{{2}^{4}}{4}\+C\right)\left(\frac{{\left(1\right)}^{4}}{4}\+C\right)\=\frac{16}{4}\frac{1}{4}\+CC\=\frac{15}{4}\+0\=\frac{15}{4}$
Finally, the two values $35\/12$ and $15\/4$ are not the same. If the graph of $f$ crosses the $x$axis in the interval of integration, its definite integral will be less than the area enclosed by this graph and the $x$axis.
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