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3 years ago

PLS HELPPP!!! I NEED THIS ANSWER

Mathematics

1 answer:

andriy [413]3 years ago

5 0

Answer:

use photo math

Step-by-step explanation:

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3 years ago

Let f(x)=x^2f ( x ) = x 2. Find the Riemann sum for ff on the interval [0,2][ 0 , 2 ], using 4 subintervals of equal width and t

sladkih [1.3K]

Answer:

$A_L=1.75$

Step-by-step explanation:

We are given:

$f(x)=x^2$

$interval = [a,b] = [0,2]$

Since $n = 4$ ⇒ $\Delta x = \frac{b-a}{n} = \frac{2-0}{4}=\frac{1}{2}$

Riemann sum is area under the function given. And it is asked to find Riemann sum for the left endpoint.

$A_L= \sum\limits^{n}_{i=1}\Delta xf(x_i) = \frac{1}{2}(0^2+(\frac{1}{2})^2+1^2+(\frac{3}{2})^2)=\frac{7}{4}=1.75$

Note:

If it will be asked to find right endpoint too,

$A_R=\sum\limits^{n}_{i=1}\Delta xf(x_i) =\frac{1}{2}((\frac{1}{2})^2+1^2+(\frac{3}{2})^2+2^2)=\frac{15}{4}=3.75$

The average of left and right endpoint Riemann sums will give approximate result of the area under $f(x)=x^2$ and it can be compared with the result of integral of the same function in the interval given.

So, $(A_R+A_L)/2 = (1.75+3.75)/2=2.25$

$\int^2_0x^2dx=x^3/3|^2_0=8/3=2.67$

Result are close but not same, since one is approximate and one is exact; however, by increasing sample rates (subintervals), closer result to the exact value can be found.

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3 years ago