Question

Find the approximate area under the curve y= sqrt{x^2+1} on the interval [0,1] using the trapezoid method and 5 subintervals.

Answer 1

Solution

5 subintervals from 0 to 1, so length of each subinterval is 0.2.

$S = s_1+s_2+s_3+s_4+s_5$

$s_1 = 0.2\cdot(\dfrac{y(0)+y(0.2)}{2}) = 0.2\cdot\dfrac{1+\sqrt{1.04}}{2} \approx0.202 \\ s_2 = 0.2\cdot(\dfrac{y(0.2)+y(0.4)}{2}) = 0.2\cdot\dfrac{\sqrt{1.04}+\sqrt{1.16}}{2} \approx0.21 \\ s_3 = 0.2\cdot(\dfrac{y(0.4)+y(0.6)}{2}) = 0.2\cdot\dfrac{\sqrt{1.16}+\sqrt{1.36}}{2} \approx0.224 \\ s_4 = 0.2\cdot(\dfrac{y(0.6)+y(0.8)}{2}) = 0.2\cdot\dfrac{\sqrt{1.36}+\sqrt{1.64}}{2} \approx0.245 \\ s_5 = 0.2\cdot(\dfrac{y(0.8)+y(1)}{2}) = 0.2\cdot\dfrac{\sqrt{1.64}+\sqrt{2}}{2} \approx0.27 \\$

$S \approx 0.202+0.21+0.224+0.245+0.27 \approx 1.15$

Answer: 1.15