Explanation
| Answer: A |
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Area between a curve and \(x-\)axis can be calculated using integration.
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| \(\begin{aligned} A&= \int_a^b y \,dx. \end{aligned}\) |
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| Based on the given diagram, |
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| \(\begin{aligned} a&=2. \\\\ b&=5. \\\\ y&=\dfrac{1}{3}x(x+2)(x-4) \end{aligned}\) |
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| Divide the shaded region into two parts \(A\) and \(B\). |
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| Area of \(\boldsymbol{ A}\) |
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\(\begin{aligned} \int_{-2}^0 y \,dx &=\int_{-2}^0 \dfrac{1}{3}x(x+2)(x-4) \,dx\\\\ &=\dfrac{1}{3} \int_{-2}^0x(x^2-2x-8) \, dx \\\\ &=\dfrac{1}{3} \int_{-2}^0 x^3-2x^2-8x \, dx \\\\ &=\dfrac{1}{3} \left[ \dfrac{x^4}{4}-\dfrac{2x^3}{3}-\dfrac{8x^2}{2} \right]_{-2}^0\\\\ &=\dfrac{1}{3} \left[ \dfrac{x^4}{4}-\dfrac{2x^3}{3}-4x^2 \right]_{-2}^0 \\\\ &=\dfrac{1}{3} \left[ \dfrac{(0)^4}{4}-\dfrac{2(0)^3}{3}-4(0) ^2\right] -\left[ \dfrac{(-2)^4}{4}-\dfrac{2(-2)^3}{3}-4(-2) ^2 \right] \\\\ &=\dfrac{1}{3} \left[ 0- \left(-\dfrac{20}{3} \right) \right] \\\\ &=\dfrac{1}{3} \left( 0 +\dfrac{20}{3} \right) \\\\ &=\dfrac{1}{3} \left( \dfrac{20}{3} \right) \\\\ &=\dfrac{20}{9}. \end{aligned}\)
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Area of \(\boldsymbol{ B}\)
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\(\begin{aligned} \int_{0}^4 y \,dx &=\int_{0}^4 \dfrac{1}{3}x(x+2)(x-4) \,dx\\\\ &=\dfrac{1}{3} \int_{-2}^0x(x^2-2x-8) \, dx \\\\ &=\dfrac{1}{3} \int_{-2}^0 x^3-2x^2-8x \, dx \\\\ &=\dfrac{1}{3} \left[ \dfrac{x^4}{4}-\dfrac{2x^3}{3}-\dfrac{8x^2}{2} \right]_{0}^4\\\\ &=\dfrac{1}{3} \left[ \dfrac{x^4}{4}-\dfrac{2x^3}{3}-4x^2 \right]_{0}^4 \\\\ &=\dfrac{1}{3} \left[ \dfrac{(4)^4}{4}-\dfrac{2(4)^3}{3}-4(4) ^2\right] -\left[ \dfrac{(0)^4}{4}-\dfrac{2(0)^3}{3}-4(0) ^2 \right] \\\\ &=\dfrac{1}{3} \left[-\dfrac{128}{3}-0\right] \\\\ &=\dfrac{1}{3} \left( -\dfrac{128}{3} \right) \\\\ &=-\dfrac{128}{9}. \end{aligned}\)
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Then, the area of \(B\) is
\(\left| -\dfrac{128}{9} \right|=\dfrac{128}{9}.\)
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Hence, the area of the shaded region is
Area of \(A\) \(+\) Area of \(B\)
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\(\dfrac{20}{9}+\dfrac{128}{9}=\dfrac{148}{9} \text{ unit}^2\)
| Jawapan: A |
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Luas di antara lengkung dengan paksi\(-x\) dapat dicari dengan menggunakan kaedah pengamiran.
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| \(\begin{aligned} L&= \int_a^b y \,dx. \end{aligned}\) |
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| Berdasarkan rajah yang diberi, |
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| \(\begin{aligned} a&=2. \\\\ b&=5. \\\\ y&=\dfrac{1}{3}x(x+2)(x-4) \end{aligned}\) |
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| Bahagikan rantau berlorek kepada dua bahagian iaitu \(A\) dan \(B\) . |
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\(\begin{aligned} \int_{-2}^0 y \,dx &=\int_{-2}^0 \dfrac{1}{3}x(x+2)(x-4) \,dx\\\\ &=\dfrac{1}{3} \int_{-2}^0x(x^2-2x-8) \, dx \\\\ &=\dfrac{1}{3} \int_{-2}^0 x^3-2x^2-8x \, dx \\\\ &=\dfrac{1}{3} \left[ \dfrac{x^4}{4}-\dfrac{2x^3}{3}-\dfrac{8x^2}{2} \right]_{-2}^0\\\\ &=\dfrac{1}{3} \left[ \dfrac{x^4}{4}-\dfrac{2x^3}{3}-4x^2 \right]_{-2}^0 \\\\ &=\dfrac{1}{3} \left[ \dfrac{(0)^4}{4}-\dfrac{2(0)^3}{3}-4(0) ^2\right] -\left[ \dfrac{(-2)^4}{4}-\dfrac{2(-2)^3}{3}-4(-2) ^2 \right] \\\\ &=\dfrac{1}{3} \left[ 0- \left(-\dfrac{20}{3} \right) \right] \\\\ &=\dfrac{1}{3} \left( 0 +\dfrac{20}{3} \right) \\\\ &=\dfrac{1}{3} \left( \dfrac{20}{3} \right) \\\\ &=\dfrac{20}{9}. \end{aligned}\)
\(\begin{aligned} \int_{0}^4 y \,dx &=\int_{0}^4 \dfrac{1}{3}x(x+2)(x-4) \,dx\\\\ &=\dfrac{1}{3} \int_{-2}^0x(x^2-2x-8) \, dx \\\\ &=\dfrac{1}{3} \int_{-2}^0 x^3-2x^2-8x \, dx \\\\ &=\dfrac{1}{3} \left[ \dfrac{x^4}{4}-\dfrac{2x^3}{3}-\dfrac{8x^2}{2} \right]_{0}^4\\\\ &=\dfrac{1}{3} \left[ \dfrac{x^4}{4}-\dfrac{2x^3}{3}-4x^2 \right]_{0}^4 \\\\ &=\dfrac{1}{3} \left[ \dfrac{(4)^4}{4}-\dfrac{2(4)^3}{3}-4(4) ^2\right] -\left[ \dfrac{(0)^4}{4}-\dfrac{2(0)^3}{3}-4(0) ^2 \right] \\\\ &=\dfrac{1}{3} \left[-\dfrac{128}{3}-0\right] \\\\ &=\dfrac{1}{3} \left( -\dfrac{128}{3} \right) \\\\ &=-\dfrac{128}{9}. \end{aligned}\)
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Maka, luas \(B\) adalah
\(\left| -\dfrac{128}{9} \right|=\dfrac{128}{9}.\)
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Maka, luas bagi rantau berlorek adalah
Luas \(A\) \(+\) Luas \(B\)
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\(\dfrac{20}{9}+\dfrac{128}{9}=\dfrac{148}{9} \text{ unit}^2\)
| Answer: A |
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Area between a curve and \(x-\)axis can be calculated using integration.
|
| |
| \(\begin{aligned} A&= \int_a^b y \,dx. \end{aligned}\) |
| |
| Based on the given diagram, |
| |
| \(\begin{aligned} a&=2. \\\\ b&=5. \\\\ y&=\dfrac{1}{3}x(x+2)(x-4) \end{aligned}\) |
| |
| Divide the shaded region into two parts \(A\) and \(B\). |
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| Area of \(\boldsymbol{ A}\) |
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\(\begin{aligned} \int_{-2}^0 y \,dx &=\int_{-2}^0 \dfrac{1}{3}x(x+2)(x-4) \,dx\\\\ &=\dfrac{1}{3} \int_{-2}^0x(x^2-2x-8) \, dx \\\\ &=\dfrac{1}{3} \int_{-2}^0 x^3-2x^2-8x \, dx \\\\ &=\dfrac{1}{3} \left[ \dfrac{x^4}{4}-\dfrac{2x^3}{3}-\dfrac{8x^2}{2} \right]_{-2}^0\\\\ &=\dfrac{1}{3} \left[ \dfrac{x^4}{4}-\dfrac{2x^3}{3}-4x^2 \right]_{-2}^0 \\\\ &=\dfrac{1}{3} \left[ \dfrac{(0)^4}{4}-\dfrac{2(0)^3}{3}-4(0) ^2\right] -\left[ \dfrac{(-2)^4}{4}-\dfrac{2(-2)^3}{3}-4(-2) ^2 \right] \\\\ &=\dfrac{1}{3} \left[ 0- \left(-\dfrac{20}{3} \right) \right] \\\\ &=\dfrac{1}{3} \left( 0 +\dfrac{20}{3} \right) \\\\ &=\dfrac{1}{3} \left( \dfrac{20}{3} \right) \\\\ &=\dfrac{20}{9}. \end{aligned}\)
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Area of \(\boldsymbol{ B}\)
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\(\begin{aligned} \int_{0}^4 y \,dx &=\int_{0}^4 \dfrac{1}{3}x(x+2)(x-4) \,dx\\\\ &=\dfrac{1}{3} \int_{-2}^0x(x^2-2x-8) \, dx \\\\ &=\dfrac{1}{3} \int_{-2}^0 x^3-2x^2-8x \, dx \\\\ &=\dfrac{1}{3} \left[ \dfrac{x^4}{4}-\dfrac{2x^3}{3}-\dfrac{8x^2}{2} \right]_{0}^4\\\\ &=\dfrac{1}{3} \left[ \dfrac{x^4}{4}-\dfrac{2x^3}{3}-4x^2 \right]_{0}^4 \\\\ &=\dfrac{1}{3} \left[ \dfrac{(4)^4}{4}-\dfrac{2(4)^3}{3}-4(4) ^2\right] -\left[ \dfrac{(0)^4}{4}-\dfrac{2(0)^3}{3}-4(0) ^2 \right] \\\\ &=\dfrac{1}{3} \left[-\dfrac{128}{3}-0\right] \\\\ &=\dfrac{1}{3} \left( -\dfrac{128}{3} \right) \\\\ &=-\dfrac{128}{9}. \end{aligned}\)
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Then, the area of \(B\) is
\(\left| -\dfrac{128}{9} \right|=\dfrac{128}{9}.\)
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Hence, the area of the shaded region is
Area of \(A\) \(+\) Area of \(B\)
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\(\dfrac{20}{9}+\dfrac{128}{9}=\dfrac{148}{9} \text{ unit}^2\)