# Evaluate Definite Integral using Right Hand Rule

• DemiMike
In summary, the integral is evaluated using the definiton of the definite inegral and the right hand rule.
DemiMike
Use the definiton of the definite inegral (with right hand rule) to evaluate the following integral. Show work please

Can NOT use shortcut method.. but be the long process

1
S (3x^2 - 5x - 6) dx
-4

$$\int_{-4}^{1}(3x^2-5x-6)dx=\int_{-4}^{1}3x^2dx-\int_{-4}^{1}5x-\int_{-4}^{1}6dx=3\frac{x^3}{3}-5\frac{x^2}{2}-6x]_{-4}^{1}$$

=$$\frac{145}{2}$$

DemiMike said:
Use the definiton of the definite inegral (with right hand rule) to evaluate the following integral. Show work please
But i don't understand "with right hand rule".What do you mean?

I suspect DemiMike means using the Riemann sum definition, using the right end point of each subinterval as the height of the rectangle.

And, of course, requiring that it be done in a particular way makes it sound an awful lot like homework, so I am moving it. DemiMike, please post school work in the appropriate place.

Also, whether homework or not, you must show some effort to do the problem yourself. This particular problem is a bit tedious but a straightforward calculation.

HallsofIvy said:
I suspect DemiMike means using the Riemann sum definition, using the right end point of each subinterval as the height of the rectangle.

And, of course, requiring that it be done in a particular way makes it sound an awful lot like homework, so I am moving it. DemiMike, please post school work in the appropriate place.

Also, whether homework or not, you must show some effort to do the problem yourself. This particular problem is a bit tedious but a straightforward calculation.

this is what i got ./.i want to see what other people get first

∫[-4,1] (3x^2 - 5x - 6) dx =
lim[n-->∞] 5/n ∑[i=1 to n] {3(-4 + 5/n)² - 5(-4 + 5/n) - 6} =
lim[n-->∞] 5*∑[i=1 to n] (48/n - 120i/n² + 75i²/n³ + 20/n -25i/n² - 6/n) =
lim[n-->∞] 5*∑[i=1 to n] (62/n - 145i/n² + 75i²/n³) =
lim[n-->∞] 5[62n/n - 145n(n+1)/(2n²) + 75n(n+1)(2n+1)/(6n³)] =
5(62 - 145/2 + 25) = 72.5

∑[i=1 to n] 1 = n
∑[i=1 to n] i = n(n+1)/2
∑[i=1 to n] i² = n(n+1)(2n+1)/6

so any 1 ?

Looks good

## What is the "Right Hand Rule" for evaluating definite integrals?

The Right Hand Rule is a method for approximating the value of a definite integral by dividing the interval into equal-width subintervals and using the right endpoint of each subinterval as the height of a rectangle. The sum of these rectangles represents an approximation of the integral, with a more accurate result achieved by using a greater number of subintervals.

## When should the "Right Hand Rule" be used for evaluating definite integrals?

The Right Hand Rule is typically used when the function being integrated is non-negative on the interval of integration, as the right endpoints of the subintervals will always give a conservative estimate of the integral's value. It is also useful for functions that are not continuous or are difficult to integrate analytically.

## What are the limitations of using the "Right Hand Rule" for evaluating definite integrals?

The Right Hand Rule is limited by the fact that it only provides an approximation of the integral, and the accuracy of the approximation depends on the number of subintervals used. It is also only applicable to functions that are non-negative on the interval of integration.

## How does the accuracy of the "Right Hand Rule" improve with a greater number of subintervals?

As the number of subintervals used in the "Right Hand Rule" increases, the width of each subinterval decreases, resulting in a closer approximation to the actual value of the integral. This is because smaller subintervals better capture the shape of the function being integrated and reduce the amount of area that is over or under-estimated.

## Are there any alternatives to the "Right Hand Rule" for evaluating definite integrals?

Yes, there are other methods for evaluating definite integrals, such as the Left Hand Rule, Midpoint Rule, and Trapezoidal Rule. Each of these methods has its own advantages and limitations and may be more suitable for certain types of functions or intervals. It is important to understand and be able to apply multiple methods in order to obtain more accurate results.

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