Mathematics, 04.09.2020 19:01 elchapo8362

Evaluate the Riemann sum for f(x) = 3 - 1/2 times x between 2 and 14 where the endpoints are included with six subintervals taking the sample points to be the left endpoints. Explain, with aid of a diagram, what the Riemann sum represents.

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Answer from: jennarosebrand3518

-6

Step-by-step explanation:

Given that :

we are to evaluate the Riemann sum for $f(x) = 3 - \dfrac{1}{2}x$ from 2 ≤ x ≤ 14

where the endpoints are included with six subintervals, taking the sample points to be the left endpoints.

The Riemann sum can be computed as follows:

$L_6 = \int ^{14}_{2}3- \dfrac{1}{2}x \dx = \lim_{n \to \infty} \sum \limits ^6 _{i=1} \ f (x_i -1) \Delta x$

where:

$\Delta x = \dfrac{b-a}{a}$

a = 2

b =14

n = 6

∴

$\Delta x = \dfrac{14-2}{6}$

$\Delta x = \dfrac{12}{6}$

$\Delta x =2$

Hence;

$x_0 = 2 \\ \\ x_1 = 2+2 =4\\ \\ x_2 = 2 + 2(2) \\ \\ x_i = 2 + 2i$

Here, we are using left end-points, then:

x_i-1 = 2+ 2(i-1)

Replacing it into Riemann equation;

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} \begin {pmatrix}3 - \dfrac{1}{2} \begin {pmatrix} 2+2 (i-1) \end {pmatrix} \end {pmatrix}2$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - (2+2(i-1))$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - (2+2i-2)$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 -2i$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 2i$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - 2 \lim_{n \to \infty} \sum \imits ^{6}_{i=1} i$

Estimating the integrals, we have :

$= 6n - 2 ( \dfrac{n(n-1)}{2})$

= 6n - n(n+1)

replacing thevalue of n = 6 (i.e the sub interval number), we have:

= 6(6) - 6(6+1)

= 36 - 36 -6

= -6

Answer from: Quest

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13

Answer from: Quest

starting with:

5 + x - 12 = 2x - 7

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)

5 - 12 = x - 7 (combine and cancel x's from both sides)

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)

0 = x (add numbers together on both sides)

so in order for the equation to be true, x must be 0.

trying x = -0.5

5 + x - 12 = 2x - 7 (original equation)

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)

5 + -0.5 - 12 = -1 - 7 (did multiplication)

-7.5 = -8 (did addition on both sides)

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.

trying x = 0

5 + x - 12 = 2x - 7 (original equation)

5 + 0 - 12 = 2(0) - 7 (plug in value of x)

5 + 0 - 12 = 0 - 7 (did multiplication)

-7 = -7 (did addition on both sides)

since -7 equals -7, that demonstrates that the value of 0 for x is correct.

trying x = 1

5 + x - 12 = 2x - 7 (original equation)

5 + 1 - 12 = 2(1) - 7 (plug in value of x)

5 + 1 - 12 = 2 - 7 (did multiplication)

-6 = -5 (did addition on both sides)

since -6 is obviously not equal to -5, the value of 1 for x is wrong.

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