Ngā Tauira Pātai me te Kōrero mō te Whakamahinga o ngā Whakaurunga i roto i te Tatau i te Horahanga o te Papa Papatahi
I roto i te whakaakoranga pāngarau, he maha ngā wā ka kitea ngā taunga whakauru i roto i te tātaitai. Ko tētahi o ngā whakamahinga rongonui o ngā taunga whakauru ko te tatau i te horahanga i raro i tētahi kōpiko, i tētahi papa rānei. Ka matapakihia e tēnei tuhinga ētahi tauira raruraru, me te matapaki i te whakamahinga o ngā taunga whakauru ki te tatau i te horahanga o tētahi papa.
Kupu Whakataki ki te Ariā
I mua i te neke atu ki te tauira raruraru, me arotake te ariā taketake o te tatau i te horahanga i raro i te kōpiko mā te whakamahi i ngā taunga whakauru. Mena he mahi f(x) tā tātou e haere tonu ana i runga i te wā [a, b], ko te horahanga i raro i te kōpiko y = f(x) mai i x = a ki x = b ko:
\[ L = \int_{a}^{b} f(x) \, dx \]
I te taha āhuahanga, ko te tikanga o tēnei kei te tāpirihia e tātou te horahanga o tētahi tapawhā angiangi rawa mai i x = a ki x = b.
Tauira Pātai 1
Pātai
Tātaihia te horahanga i raro i te kōpiko y = x² i te wā [1, 3].
Kōrero
Hei tatau i te horahanga, ka whakamahia e mātou te taupū:
\[ L = \int_{1}^{3} x^2 \, dx \]
Ka tīmata tātou mā te kimi i te ātete-whakaputa o \( x^2 \). Ko te ātete-whakaputa o \( x^2 \) ko \( \frac{x^3}{3} \). Kātahi ka noho te taupū hei:
\[ L = \left[ \frac{x^3}{3} \right]_{1}^{3} \]
Kia maumahara me aromatawai tātou i te taupatupatu i runga i ngā rohe o te taupū:
\[ L = \left( \frac{3^3}{3} \right) – \left( \frac{1^3}{3} \right) \]
\[ L = \left( \frac{27}{3} \right) – \left( \frac{1}{3} \right) \]
\[ L = 9 – \frac{1}{3} \]
\[ L = \frac{27}{3} – \frac{1}{3} \]
\[ L = \frac{26}{3} \]
Nō reira, ko te horahanga i raro i te kōpiko y = x² mai i x = 1 ki x = 3 ko:
\[ \frac{26}{3} \, \text{wāhanga rohe} \]
Tauira Pātai 2
Pātai
Tāutuhia te horahanga o te rohe e herea ana e te kōpiko y = x³ me ngā rārangi x = 1 me x = 2.
Kōrero
Hei tatau i te horahanga, ka whakamahia e mātou te taupū:
\[ L = \int_{1}^{2} x^3 \, dx \]
Pērā i ngā wā katoa, ka tīmata tātou mā te kimi i te ātete-whakaputa o \( x^3 \). Ko te ātete-whakaputa o \( x^3 \) ko \( \frac{x^4}{4} \). Ka noho te taupū hei:
\[ L = \left[ \frac{x^4}{4} \right]_{1}^{2} \]
Arotakehia ngā rohe o te taupū:
\[ L = \left( \frac{2^4}{4} \right) – \left( \frac{1^4}{4} \right) \]
\[ L = \left( \frac{16}{4} \right) – \left( \frac{1}{4} \right) \]
\[ L = 4 – \frac{1}{4} \]
\[ L = \frac{16}{4} – \frac{1}{4} \]
\[ L = \frac{15}{4} \]
Nō reira, ko te horahanga i raro i te kōpiko y = x³ mai i x = 1 ki x = 2 ko:
\[ \frac{15}{4} \, \text{wāhanga rohe} \]
Tauira Pātai 3
Pātai
Tāutuhia te horahanga o te rohe e herea ana e ngā kōpiko y = x² + 1 me y = 2x + 2 i te wā x = 0 ki x = 1.
Kōrero
Tuatahi, me kimi tātou i ngā pūwāhi whakawhiti hei whakatau i ngā rohe o te whakaurunga. Ko te otinga ki \( x^2 + 1 = 2x + 2 \):
\[ x^2 + 1 = 2x + 2 \]
\[ x^2 – 2x – 1 = 0 \]
Mā te whakamahi i te tātai tapawhā:
\[ x = \frac{2 \pm \sqrt{4 + 4}}{2} \]
\[ x = \frac{2 \pm \sqrt{8}}{2} \]
\[ x = \frac{2 \pm 2\sqrt{2}}{2} \]
\[ x = 1 \pm \sqrt{2} \]
Heoi, mō ngā rohe o runga me o raro i waenga i te 0 me te 1, kāore e hiahiatia te whakamahi i te otinga tapawhā, engari ko te rohe taupū noa mai i te 0 ki te 1. Muri iho, tatauhia te horahanga o te pihi y o runga me te tango i te pihi y o raro e ai ki ēnei rohe:
\[ L = \int_{0}^{1} [(2x + 2) – (x^2 + 1)] \, dx \]
Te whakangawari i te mahi:
\[ L = \int_{0}^{1} (2x + 2 – x^2 – 1) \, dx \]
\[ L = \int_{0}^{1} (-x^2 + 2x + 1) \, dx \]
Muri iho, ka kitea e tātou te antiderivative:
Ko te ātete-whakaputa o \( (-x^2) \) ko \( -\frac{x^3}{3} \),
Ko te ātete-whakaputa o \( (2x) \) ko \( x^2 \),
Ko te ātete-whakaputa o \( (1) \) ko \( x \).
Nō reira,
\[ L = \left. \left(-\frac{x^3}{3} + x^2 + x \right) \right|_0^1 \]
Aromatawai e whai ake nei:
\[ L = \left[ -\frac{1^3}{3} + 1^2 + 1 \matau] – \left[ -\frac{0^3}{3} + 0^2 + 0 \matau] \]
\[ L = \left[ -\frac{1}{3} + 1 + 1 \matau] – \left[ 0 \matau] \]
\[ L = -\frac{1}{3} + 2 \]
\[ L = \frac{6}{3} – \frac{1}{3} \]
\[ L = \frac{5}{3} \]
Nō reira, ko te horahanga o te rohe e herea ana e ngā kōpiko y = x² + 1 me y = 2x + 2 i te mokowā [0, 1] ko:
\[ \frac{5}{3} \, \text{wāhanga rohe} \]
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Mai i ngā tauira i runga ake nei, ka kitea e tātou te whakamahinga o ngā taunga whakauru hei tatau i te horahanga i raro i tētahi kōpiko, i waenganui rānei i ngā kōpiko e rua. Mā te mārama tika ki ngā ariā taketake o ngā taunga whakauru me ngā tikanga ārai-whakaputa, ka tino whai hua te tatau i ēnei horahanga. Ko te tumanako, kua whakanuia e tēnei tuhinga tō tātou māramatanga ki te whakamahinga o ngā taunga whakauru i te ao tūturu, inā koa i te mara o te ine i te horahanga o ngā mata papatahi.
