Estimate the area under the graph of f(x)=2x^2-12x+22 over the interval [0,2] using four approximating rectangles and right endpoints.

Estimate the area under the graph of fx2x212x22 over the interval 02 using four approximating rectangles and right endpoints class=

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Answer:

The right Riemann sum is 21.5.

The left Riemann sum is 29.5.

Step-by-step explanation:

The right Riemann sum (also known as the right endpoint approximation) uses the right endpoints of a sub-interval:

[tex]\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f(x_1)+f(x_2)+f(x_3)+...+f(x_{n-1})+f(x_{n})\right)[/tex], where [tex]\Delta{x}=\frac{b-a}{n}[/tex].

To find the Riemann sum for [tex]\int_{0}^{2}\left(2 x^{2} - 12 x + 22\right)\ dx[/tex] with 4 rectangles, using right endpoints you must:

We have that a = 0, b = 2, n = 4. Therefore, [tex]\Delta{x}=\frac{2-0}{4}=\frac{1}{2}[/tex].

Divide the interval [0,2] into n = 4 sub-intervals of length [tex]\Delta{x}=\frac{1}{2}[/tex]:

[tex]\left[0, \frac{1}{2}\right], \left[\frac{1}{2}, 1\right], \left[1, \frac{3}{2}\right], \left[\frac{3}{2}, 2\right][/tex]

Now, we just evaluate the function at the right endpoints:

[tex]f\left(x_{1}\right)=f\left(\frac{1}{2}\right)=\frac{33}{2}=16.5\\\\f\left(x_{2}\right)=f\left(1\right)=12\\\\f\left(x_{3}\right)=f\left(\frac{3}{2}\right)=\frac{17}{2}=8.5\\\\f\left(x_{4}\right)=f(b)=f\left(2\right)=6[/tex]

Finally, just sum up the above values and multiply by [tex]\Delta{x}=\frac{1}{2}[/tex]:

[tex]\frac{1}{2}(16.5+12+8.5+6)=21.5[/tex]

The left Riemann sum (also known as the left endpoint approximation) uses the left endpoints of a sub-interval:

[tex]\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f(x_0)+f(x_1)+2f(x_2)+...+f(x_{n-2})+f(x_{n-1})\right)[/tex], where [tex]\Delta{x}=\frac{b-a}{n}[/tex].

To find the Riemann sum for [tex]\int_{0}^{2}\left(2 x^{2} - 12 x + 22\right)\ dx[/tex] with 4 rectangles, using left endpoints you must:

We have that a = 0, b = 2, n = 4. Therefore, [tex]\Delta{x}=\frac{2-0}{4}=\frac{1}{2}[/tex].

Divide the interval [0,2] into n = 4 sub-intervals of length [tex]\Delta{x}=\frac{1}{2}[/tex]:

[tex]\left[0, \frac{1}{2}\right], \left[\frac{1}{2}, 1\right], \left[1, \frac{3}{2}\right], \left[\frac{3}{2}, 2\right][/tex]

Now, we just evaluate the function at the left endpoints:

[tex]f\left(x_{0}\right)=f(a)=f\left(0\right)=22\\\\f\left(x_{1}\right)=f\left(\frac{1}{2}\right)=\frac{33}{2}=16.5\\\\f\left(x_{2}\right)=f\left(1\\\right)=12\\\\f\left(x_{3}\right)=f\left(\frac{3}{2}\right)=\frac{17}{2}=8.5[/tex]

Finally, just sum up the above values and multiply by [tex]\Delta{x}=\frac{1}{2}[/tex]:

[tex]\frac{1}{2}(22+16.5+12+8.5)=29.5[/tex]