Find the area of the region enclose by the curve y = 1 − x^2 , and the line y = 1 + 3x, x = 1.


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Question Number 109674 by I want to learn more last updated on 24/Aug/20

$Find the area of the region enclose by the curve y = 1 - x^{2},$ $and the line y = 1 + 3 x, x = 1 .$
	Answered by bobhans last updated on 25/Aug/20

	$\frac{\iddots ♭ o ♭ \iddots}{h a n s}$ $A r e a 1 = \underset{0}{\int^{1}} (1 + 3 x) - (1 - x^{2}) d x$ $= \underset{0}{\int^{1}} (3 x + x^{2}) d x = {[\frac{3}{2} x^{2} + \frac{1}{3} x^{3}]}_{0}^{1}$ $= \frac{3}{2} + \frac{1}{3} = \frac{11}{6}$ $A r e a 2 = \frac{Δ \sqrt{Δ}}{6 a^{2}} = \frac{9 \sqrt{9}}{6.1} = \frac{9.3}{6} = \frac{27}{6}$ $w h e r e 1 + 3 x = 1 - x^{2} \Rightarrow x^{2} + 3 x = 0; Δ = 9 - 4.1.0 = 9$ $T o t a l l y a r e a = \frac{11 + 27}{6} = \frac{38}{6}$
	Commented by I want to learn more last updated on 25/Aug/20

	$Thanks sir$
	Answered by 1549442205PVT last updated on 25/Aug/20
	$We need find intesection point of two graphs of the functions y=1−x^2 and y=3x+1. { ((y=1−x^2 )),((y=3x+1)) :}⇔ { ((1−x^2 =3x+1(1))),((y=3x+1)) :} (1)⇔x^2 +3x=0⇔x(x+3)=0⇔x∈{0;−3} y∈{1;−8}⇒A(0;1),B(−3;−8) S_1 =∫_0 ^( 1) ∣3x+1−(1−x^2 )∣dx=∫_0 ^1 (x^2 +3x)dx S_2 =∫_(−3) ^( 0) ∣(x^2 +3x)∣dx=−∫_(−3) ^( 0) (x^2 +3x)∣dx −(x^3 /3)−((3x^2 )/2)=−9+((27)/2)=(9/2) =((x^3 /3)+((3x^2 )/2))_0 ^1 =(1/3)+(3/2)=((11)/6) .From that S=S_1 +S_2 =(9/2)+((11)/6)=((19)/3)$
	$We need find intesection point of two$ $graphs of the functions y = 1 - x^{2} and$ $y = 3 x + 1 .$ ${\begin{cases} y = 1 - x^{2} \\ y = 3 x + 1 \end{cases} \Leftrightarrow {\begin{cases} 1 - x^{2} = 3 x + 1 (1) \\ y = 3 x + 1 \end{cases}$ $(1) \Leftrightarrow x^{2} + 3 x = 0 \Leftrightarrow x (x + 3) = 0 \Leftrightarrow x \in {0; - 3}$ $y \in {1; - 8} \Rightarrow A (0; 1), B (- 3; - 8)$ $S_{1} = \int_{0}^{1} ∣ 3 x + 1 - (1 - x^{2}) ∣ dx = \int_{0}^{1} (x^{2} + 3 x) dx$ $S_{2} = \int_{- 3}^{0} ∣ (x^{2} + 3 x) ∣ dx = - \int_{- 3}^{0} (x^{2} + 3 x) ∣ dx$ $- \frac{x^{3}}{3} - \frac{{3 x}^{2}}{2} = - 9 + \frac{27}{2} = \frac{9}{2}$ $= {(\frac{x^{3}}{3} + \frac{3 x^{2}}{2})}_{0}^{1} = \frac{1}{3} + \frac{3}{2} = \frac{11}{6} . From that$ $S = S_{1} + S_{2} = \frac{9}{2} + \frac{11}{6} = \frac{19}{3}$
	Commented by I want to learn more last updated on 25/Aug/20

	$Thanks sir$
	Commented by 1549442205PVT last updated on 25/Aug/20

	$You are welcome!$
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