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Aromātai
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\int x^{2}+1\mathrm{d}x
Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.
\int x^{2}\mathrm{d}x+\int 1\mathrm{d}x
Kōmitimititia te kīanga tapeke mā te kīanga.
\frac{x^{3}}{3}+\int 1\mathrm{d}x
Nā te mea \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int x^{2}\mathrm{d}x ki te \frac{x^{3}}{3}.
\frac{x^{3}}{3}+x
Kimihia te tau tōpū o 1 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.
\frac{1^{3}}{3}+1-\left(\frac{0^{3}}{3}+0\right)
Ko te tau tōpū tautuhi ko te pārōnaki kōaro o te kīanga i aromātaitia i te tepe tōrunga o te pāwhaitua, tangohia te pārōnaki kōaro i aromātaitia i te tepe tōraro o te pāwhaitua.
\frac{4}{3}
Whakarūnātia.