# calculate integral `int_-3^4(4x-x^2) dx` using Riemann Sum and regular partition with n sub intervals.

*print*Print*list*Cite

### 1 Answer

You should remember what is the formula taht helps you to evaluate the definite integral as a Riemann sum such that:

`int_-3^4 (4x - x^2)dx = int_-3^4 4x dx - int_-3^4 x^2 dx = lim_(n->oo) sum_(i=1)^n f(c_i)*Delta x_i`

You should split the interval `[-3,4]` into n subdivisions of equal lengths such that:

`Delta = (4 - (-3))/n = 7/n`

Using the right endpoints `c_i` of each interval yields:

`c_i = -3 + (7/n)*i`

You need to evaluate `f(c_i) ` for each function `y = 4x ` and `y = x^2` such that:

`f(c_i) = 4(-3 + (7/n)*i)`

`f(c_i) = (-3 + (7/n)*i)^2`

`int_-3^4 (4x - x^2)dx = lim_(n->oo) sum_(i=1)^n 4(-3 + (7/n)*i)*(7/n) - lim_(n->oo) sum_(i=1)^n (-3 + (7/n)*i)^2*(7/n)`

`int_-3^4 (4x - x^2)dx = lim_(n->oo) (7/n)(-12n + (7/n)(n(n+1)/2))- lim_(n->oo) (7/n)((7/n)^2*(n(n+1)(2n+1))/6 - 42/n(n(n+1))/2) + 9)`

`int_-3^4 (4x - x^2)dx = lim_(n->oo) (-12*7 + 49/2) - lim_(n->oo) (14*49/6 - 21*7 + 63/n)`

`int_-3^4 (4x - x^2)dx = -59.5- 114.3 + 147 = -26.8`

**Hence, evaluating the integral using Riemann sum yields** `int_-3^4 (4x - x^2)dx ~~ -26.8`