# How do you estimate the area under the graph of f(x) = sqrt x from x=0 to x=4 using four approximating rectangles and right endpoints?

Mar 24, 2015

For this problem: $f \left(x\right) = \sqrt{x}$

$a = 0$ and $b = 4$

the number of rectangles $= n = 4$

$\Delta x =$ the length of each subinterval = the length of each base

$\Delta x = \frac{b - a}{n} = \frac{4 - 0}{4} = 1$

To find all of the endpoints of subintervals, start at $a$ and successively add $\Delta x$ until you reach $b$

All endpoints: $0 , 1 , 2 , 3 , 4$.

The right endpoints are: $1 , 2 , 3 , 4$.

The heights at the right endpoints are:

$f \left(1\right) = \sqrt{1} = 1$
$f \left(2\right) = \sqrt{2}$
$f \left(3\right) = \sqrt{3}$
$f \left(4\right) = \sqrt{4} = 2$

Call the areas of the rectangles ${R}_{1} , {R}_{2}$ etc: Each has area $\text{base" xx "height}$. Every base is $\Delta x$ and the heights are above, so

Then the approximation we want is:
${R}_{1} + {R}_{2} + {R}_{3} + {R}_{4}$

$= \Delta x \cdot f \left(1\right) + \Delta x \cdot f \left(2\right) + \Delta x \cdot f \left(3\right) + \Delta x \cdot f \left(4\right)$

$= 1 \cdot 1 + 1 \cdot \sqrt{2} + 1 \cdot \sqrt{3} + 1 \cdot 2 = 1 + \sqrt{2} + \sqrt{3} + 2 = 3 + \sqrt{2} + \sqrt{3}$

$3 + \sqrt{2} + \sqrt{3} \approx 3 + 1.414 + 1.732 = 6.146$