Question

Approximate the area under the curve y = x^2 from x = 0 to x = 3 using a Right Endpoint approximation with 6 subdivisions.


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Answer 1

Subdivide the interval [0, 3] into 6 subintervals of equal length, ∆x = (3 - 0)/6 = 1/2. Then the partition is

$\left[0,\dfrac12\right] \cup \left[\dfrac12,1\right] \cup \left[1,\dfrac32\right] \cup \cdots \cup \left[\dfrac52,3\right]$

The right endpoint of the $i$-th subinterval is

$r_i = \dfrac12 + (i-1)\times\dfrac12 = \dfrac i2$

with $i\in\{1,2,3,\ldots,6\}$.

Then the area under the $y=x^2$ on the interval [0, 3] is approximately

$\displaystyle \int_0^3 x^2 \, dx \approx \sum_{i=1}^6 (r_i)^2 \Delta x = \frac18 \sum_{i=1}^6 i^2$

Recall that

$\displaystyle \sum_{n=1}^N n^2 = \frac{N(N+1)(2N+1)}6$

Then the approximate value of the area is

$\displaystyle \int_0^3 x^2 \, dx \approx \frac{6\times7\times13}{48} = \boxed{\frac{91}8}$

The actual value of the area is 9, so this approximation is an overestimation, as is always the case when using a right endpoint sum for an increasing function.