Find-the-area-of-the-region-enclose-by-the-curve-y-1-x-2-and-the-line-y-1-3x-x-1-

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 = 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!