Question

Compute the lower Riemann sum for the given function f(x)=x2 over the interval x∈[−1,1] with respect to the partition P=[−1,− 1 2 , 1 2 , 3 4 ,1].

Answer 1

Answer:

21/64

Step-by-step explanation:

First, we need to note that the function f(x) = x² is increasing on (0, +∞), and it is decreasing on (-∞,0)

The first interval generated by the partition is [-1, -1/2], since f is decreasing for negative values, we have that f takes its minimum values at the right extreme of the interval, hence -1/2.

The second interval is [-1/2, 1/2]. Here f takes its minimum value at 0, because f(0) = 0, and f is positive otherwise.

Since f is increasing for positive values of x, then, on the remaining 2 intervals, f takes its minimum value at their respective left extremes, in other words, 1/2 and 3/4 respectively.

We obtain the lower Riemman sum by multiplying this values evaluated in f by the lenght of their respective intervals and summing the results, thus

LP(f) = f(-1/2) * ((-1/2) - (-1)) + f(0) * (1/2 - (-1/2)) + f(1/2)* (3/4 - 1/2) + f(3/4) * (1- 3/4)

= 1/4 * 1/2 + 0 * 1 + 1/4 * 1/4 + 9/16 * 1/4 = 1/8 + 0 + 1/16 + 9/64 = 21/64

As a result, the lower Riemann sum on the partition P is 21/64

Answer 2

Answer:

The lower Reimann sum for the given function over the interval is  $\dfrac{21}{64}$.

Step-by-step explanation:

The given function is $f(x)=x^2$.

The given function is a parabolic function which is increasing in the interval 0 to infinity and decreasing in the interval negative infinity to 0.

The function will have a minimum value at $x=0$ which is $f(0)=0$.

Now, the partition P is $P=[-1,\dfrac{-1}{2}, \dfrac{1}{2}, \dfrac{3}{4}, 1]$.

The first interval from the partition is $[-1,\dfrac{-1}{2}]$. Now, the function is decreasing for the negative values of $x$. So, the minimum value of the function in this interval will be at $x=\dfrac{-1}{2}$ which is $f(\dfrac{-1}{2})=\dfrac{1}{4}$.

The second interval from the partition is $[\dfrac{-1}{2},\dfrac{1}{2}]$. Now, the function is decreasing for the negative values of $x$ and increasing for positive values of $x$. So, the minimum value of the function in this interval will be at $x=0$ which is $f(0)=0$.

The third interval from the partition is $[\dfrac{1}{2}, \dfrac{3}{4}]$. Now, the function is increasing for the positive values of $x$. So, the minimum value of the function in this interval will be at $x=\dfrac{1}{2}$ which is $f(\dfrac{1}{2})=\dfrac{1}{4}$.

The fourth interval from the partition is $[\dfrac{3}{4},1]$. Now, the function is increasing for the positive values of $x$. So, the minimum value of the function in this interval will be at $x=\dfrac{3}{4}$ which is $f(\dfrac{3}{4})=\dfrac{9}{16}$.

Thus, the lower Reimann sum will be calculated as,

$LP(f)=f(\dfrac{-1}{2})\times (\dfrac{-1}{2}-(-1))+f(0)\times (\dfrac{1}{2}-(\dfrac{-1}{2}))+f(\dfrac{1}{2})\times (\dfrac{3}{4}-\dfrac{1}{2})+f(\dfrac{3}{4})\times (1-\dfrac{3}{4})\\LP(f)=\dfrac{1}{4}\times\dfrac{1}{2}+0\times 1+\dfrac{1}{4}\times\dfrac{1}{4}+\dfrac{9}{16}\times \dfrac{1}{4}\\LP(f)=\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{9}{64}\\LP(f)=\dfrac{8+4+9}{64}\\LP(f)=\dfrac{21}{64}$

Therefore, the lower Reimann sum for the given function over the interval is  $\dfrac{21}{64}$.

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