Evaluate Definite Integral using 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

 

[coki2000]

\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}

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?

 

[HallsofIvy]

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.

 

[DemiMike]

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

 

[DemiMike]

so any 1 ?

 

[Redbelly98]

Looks good