determine-the-area-of-the-region-bounded-by-y-2x-6-0-5-and-y-x-1-

Question Number 116738 by Eric002 last updated on 06/Oct/20

d e t e r m i n e t h e a r e a o f t h e r e g i o n b o u n d e d

b y y = {(2 x + 6)}^{0.5} a n d y = x - 1

Commented by bobhans last updated on 06/Oct/20

Answered by 1549442205PVT last updated on 06/Oct/20

Question should be : Calculate the area

of region bounded by curve y = \sqrt{2 x + 6}

, the line y = x - 1 and the Ox

We find the intersection of line (d) :

y = x - 1 and the curve C : y = \sqrt{2 x + 6}

x - 1 = \sqrt{2 x + 6} \Leftrightarrow 2 x + 6 = x^{2} - 2 x + 1 (x ⩾ 1)

\Leftrightarrow x^{2} - 4 x - 5 = 0 \Leftrightarrow (x + 1) (x - 5) = 0

\Leftrightarrow x = 5 \Rightarrow y = 4 . \Rightarrow C \cap d = B (5; 4)

C \cap Ox = A (- 3; 0), d \cap Ox = C (1; 0) . Hence

The area of region bounded by C, d

and the asix Ox equal to

S = \int_{- 3}^{5} \sqrt{2 x + 6} dx - \int_{1}^{5} (x - 1) dx

Put \sqrt{2 x + 6} = u \Rightarrow 2 x + 6 = u^{2}

\Rightarrow dx = udu \int \sqrt{2 x + 6} dx = \int u^{2} du = \frac{u^{3}}{3}

\Rightarrow S = \frac{(2 x + 6) \sqrt{2 x + 6}}{3} ∣_{- 3}^{5} - {[\frac{x^{2}}{2} - x]}_{1}^{5} =

= \frac{64}{3} - 7.5 - 0.5 = \frac{64}{3} - 8 = \frac{40}{3}

S = \frac{40}{3}