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Question Number 206571 by mr W last updated on 19/Apr/24

Commented by MathematicalUser2357 last updated on 26/Dec/24
Translation - What is the area S? (This question asks to calculate the area of S)
	Answered by cortano21 last updated on 19/Apr/24

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	Commented by mr W last updated on 19/Apr/24
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	Answered by MM42 last updated on 19/Apr/24

	$f (x) = 2 x^{2}; A (a, 2 a^{2}), f^{'} (a) = 4 a$ $g (x) = 2 x^{2} - 8 x + 16; B (b, 2 b^{2} - 8 b + 16), g^{'} (b) = 4 b - 8$ $f^{'} (a) = g^{'} (b) \Rightarrow a = b - 2 (i)$ $m_{A B} = \frac{2 b^{2} - 8 b + 16 - 2 a^{2}}{b - a} = 4 a (i i)$ $(i), (i i) \Rightarrow a = 1 & b = 3$ $\Rightarrow m_{A B} = 4 \Rightarrow L : y = 4 x - 2$ $f = g \Rightarrow x = 2$ $s_{1} = \int_{1}^{2} (2 x^{2} - 4 x + 2) d x = \frac{2}{3}$ $s_{2} = \int_{2}^{3} (2 x^{2} - 8 x + 16 - 4 x + 2) d x = \frac{2}{3}$ $\Rightarrow s = \frac{4}{3} ✓$
	Commented by mr W last updated on 19/Apr/24
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	Answered by MathematicalUser2357 last updated on 28/Apr/24

	Commented by MathematicalUser2357 last updated on 26/Dec/24
	$L_A T^E X Version here (Also fixed mistakes) ↓ Q.206571 - What is the area S? min{2x^2 }=0 so start point is (0,0) min{2x^2 −8x+16}=8 so end point is (4,8) Slope: 4, Y-intercept: 0. ⇒ y=4x ← There was a mistake! must be y=4x−2. These two parabolas meet at x=2 So, S=∫_1 ^2 2x^2 dx+∫_2 ^3 (2x^2 −8x+16)dx−∫_1 ^3 (4x−2)dx =[(2/3)x^3 ]_1 ^2 +[(2/3)x^3 −4x^2 +16x]_2 ^3 −[2x^2 −2x]_1 ^3 Note: I′m using built-in Complex number calculator to find area. (to prevent negative number mistake) (((2/3)×2^3 )−((2/3)×1^3 ))+(((2/3)×3^3 −4×3^2 +16×3)−((2/3)×2^3 −4×2^2 +16×2))−((2×3^2 −2×3)−(2×1^2 −2×1)) 1.333333 So S=(4/3).$
	$L_{A} T^{E} X Version here (Also fixed mistakes) ↓$ $Q . 206571 - What is the area S ?$ $\min {2 x^{2}} = 0 so start point is (0, 0)$ $\min {2 x^{2} - 8 x + 16} = 8 so end point is (4, 8)$ $Slope : 4, Y - intercept : 0 . \Rightarrow y = 4 x \leftarrow There was a mistake! must be y = 4 x - 2 . These two parabolas meet at x = 2$ $So, S = \int_{1}^{2} 2 x^{2} d x + \int_{2}^{3} (2 x^{2} - 8 x + 16) d x - \int_{1}^{3} (4 x - 2) d x$ $= {[\frac{2}{3} x^{3}]}_{1}^{2} + {[\frac{2}{3} x^{3} - 4 x^{2} + 16 x]}_{2}^{3} - {[2 x^{2} - 2 x]}_{1}^{3}$ $Note : I^{'} m using built - in Complex number calculator$ $to find area . (to prevent negative number mistake)$ $((\frac{2}{3} \times 2^{3}) - (\frac{2}{3} \times 1^{3})) + ((\frac{2}{3} \times 3^{3} - 4 \times 3^{2} + 16 \times 3) - (\frac{2}{3} \times 2^{3} - 4 \times 2^{2} + 16 \times 2)) - ((2 \times 3^{2} - 2 \times 3) - (2 \times 1^{2} - 2 \times 1))$ $1.333333$ $So S = \frac{4}{3} .$
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