The area is given exactly by the definite integral,
![\displaystyle\int_(-3)^2(x^2+4)\,\mathrm dx=\left(\frac{x^3}3+5x\right)\bigg|_(-3)^2=\frac{95}3\approx31.67](https://img.qammunity.org/2022/formulas/mathematics/college/c23gn0pre7m1rgl7chmlgflb40vinfcadq.png)
We can write this as a Riemann sum, i.e. the infinite sum of rectangular areas:
• Split up the integration interval into n equally-spaced subintervals, each with length (2 - (-3))/n = 5/n - - this will be the width of each rectangle. The intervals would then be
[-3, -3 + 5/n], [-3 + 5/n, -3 + 10/n], …, [-3 + 5(n - 1)/n, 2]
• Over each subinterval, take the function value at some point x * to be the height.
Then the area is given by
![\displaystyle\lim_(n\to\infty)\sum_(k=1)^nf(x^*)\Delta x_k=\lim_(n\to\infty)\sum_(k=1)^nf(x^*)\frac5n](https://img.qammunity.org/2022/formulas/mathematics/college/9z2im4kq77s28ettgybkct1cte27t8xa19.png)
Now, if we take x * to be the left endpoint of each subinterval, we have
x * = -3 + 5(k - 1)/n → f (x *) = (-3 + 5(k - 1)/n)² + 4
If we instead take x * to be the right endpoint, then
x * = -3 + 5k/n → f (x *) = (-3 + 5k/n)² + 4
So as a Riemann sum, the area is represented by
![\displaystyle\lim_(n\to\infty)\sum_(k=1)^n\left(\left(-3+\frac{5k}n\right)^2+4\right)\frac5n](https://img.qammunity.org/2022/formulas/mathematics/college/fvimg3g3vfq0tsfaqlo4swsxgk1cauebpn.png)
and if you expand the summand, this is the same as
![\displaystyle\lim_(n\to\infty)\sum_(k=1)^n\left(13-\frac{30k}n+(25k^2)/(n^2)\right)\frac5n=\lim_(n\to\infty)\sum_(k=1)^n\left(\frac{65}n-(150k)/(n^2)+(125k^2)/(n^3)\right)](https://img.qammunity.org/2022/formulas/mathematics/college/hqkivilfjkwh6kw67t3yikigjkww3itb1t.png)
So from the given choices, the correct ones are
• row 1, column 1
• row 2, column 2
• row 4, column 2