How do you find Find the Riemann sum that approximates the integral #int_0^9sqrt(1+x^2)dx# using left endpoints with #n=6#?

Question

William Abdalla · Accepted Answer

Let #f(x)=sqrt{1+x^2}#. #Delta x={b-a}/n={9-0}/6=1.5# #L_6=[f(x_0)+f(x_1)+cdots+f(x_5)]cdot Delta x# #=[f(0)+f(1.5)+f(3)+f(4.5)+f(6)+f(7.5)]cdot(1.5)# #approx 34.84#

Aurora Alvarado · Answer

undefined To find the Riemann sum that approximates the integral $ \int_{0}^{9} \sqrt{1 + x^2} \, dx $ using left endpoints with $ n = 6 $, you need to partition the interval $[0, 9]$ into $ n $ subintervals of equal width and then evaluate the function at the left endpoint of each subinterval.

The width of each subinterval, $ \Delta x $, is given by:

$$ \Delta x = \frac{9 - 0}{6} = \frac{9}{6} = \frac{3}{2} $$

So, the six subintervals are:

$$ [0, \frac{3}{2}], [\frac{3}{2}, 3], [3, \frac{9}{2}], [\frac{9}{2}, 6], [6, \frac{15}{2}], [\frac{15}{2}, 9] $$

Now, evaluate the function at the left endpoint of each subinterval:

$$ f(0) = \sqrt{1 + (0)^2} = \sqrt{1} = 1 $$
$$ f\left(\frac{3}{2}\right) = \sqrt{1 + \left(\frac{3}{2}\right)^2} = \sqrt{\frac{13}{4}} $$
$$ f(3) = \sqrt{1 + 3^2} = \sqrt{10} $$
$$ f\left(\frac{9}{2}\right) = \sqrt{1 + \left(\frac{9}{2}\right)^2} = \sqrt{\frac{85}{4}} $$
$$ f(6) = \sqrt{1 + 6^2} = \sqrt{37} $$
$$ f\left(\frac{15}{2}\right) = \sqrt{1 + \left(\frac{15}{2}\right)^2} = \sqrt{\frac{221}{4}} $$

Finally, compute the Riemann sum:

$$ \text{Riemann Sum} = \Delta x \left( f(0) + f\left(\frac{3}{2}\right) + f(3) + f\left(\frac{9}{2}\right) + f(6) + f\left(\frac{15}{2}\right) \right) $$