Answer to Question #174482 in Calculus for Sakau

Question

(a) Find the area enclosed by the curve 𝑦 = 25 − 𝑥2

and the straight line 

𝑦 = 𝑥 + 13 correct to 1 decimal place.

(b)Calculate the arc length of the curve 𝑦 = 𝑥

3

2 between 𝑥 = 0 𝑎𝑛𝑑 𝑥 = 1

correct to 1 decimal place

Answer 1

To get the points of intersection,

y=y

$25-x^2=x+13\\ x^2+x-12=0\\ x=-4, x=3$

The 2 points of intersection are at x = -4 and x = 3.

$y=25- x^2\\ INT(x) = 25x - {\frac{x^{3}}{3}}\\ INT(3) = 75 - 9 = 66\\ INT(x) = 25x - {\frac{x^{3}}{3}}\\ INT(-4) = -100 + \frac{64}{3} = \frac{-236}{3}$

Area under parabola to x-axis,

$= 66 +\frac{236}{3} = \frac{434}{3}\\ \text{Area under line}= 87.5 sq unit.\\ b)\\ F(x)=x^{\frac{3}{2}}\\ (f'(x))^2=9x\\ \text{Arc length}=\int_0^1 \sqrt{1+9x}dx\\ u=1+9x, du=9dx\\ When, x=0, u=1, When, x=1, u=10\\ ={\frac{1}{9}}\int_0^1\sqrt{1+9x}dx\\ ={\frac{1}{9}}\int_1^{10}\sqrt{u}du\\ ={\frac{1}{9}}\times{\frac{2}{3}} u^{\frac{3}{2}}|_1^{10}\\ =2.3units$