# Consider the following.f(x) = 49 - x^{2} from x = 1 to x = 7, 4 subintervals(a) Approximate the

Consider the following.
$f\left(x\right)=49-{x}^{2}$
from subintervals
(a) Approximate the area under the curve over the specified interval by using the indicated number of subintervals (or rectangles) and evaluating the function at the right-hand endpoints of the subintervals.
(b) Approximate the area under the curve by evaluating the function at the left-hand endpoints of the subintervals.

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mhalmantus

(a) Given that, the function is $f\left(x\right)=49-{x}^{2}$ on the interval $\left[1,7\right],n=4.$
It is known that, the formula for the Riemann-sum is ${\int }_{a}^{b}f\left(x\right)dx=\mathrm{\Delta }x\sum _{i=0}^{\mathrm{\infty }}f\left(xi\right),$ where ${\mathrm{\Delta }}_{x}=\frac{b-a}{n}.$
Obtain the value of ${\mathrm{\Delta }}_{x}=\frac{b-a}{n}$
$\left(a=1,b=7,n=4\right)$
${\mathrm{\Delta }}_{x}=\frac{7-1}{4}$
$=\frac{6}{4}$
$=\frac{3}{2}$
Divide the interval $\left[1,7\right]$ into $n=4$ sub-intervals with length ${\mathrm{\Delta }}_{x}=\frac{3}{2}$ as $\left[1,\frac{5}{2}\right],\left[\frac{5}{2},4\right]\left[4,\frac{11}{2}\right],\left[\frac{11}{2},7\right],\left[\frac{11}{2},7\right]$ and the right end-points are ${x}_{1}=\frac{5}{2},{x}_{2}=4,{x}_{3}=\frac{11}{2},{x}_{4}=7$. Find the area under the curve over $\left[1,7\right]$ by using the right-end points as follows,
${\int }_{1}^{7}\left(49-{x}^{2}\right)dx\approx \mathrm{\Delta }x\left(f\left({x}_{1}\right)+f\left({x}_{2}\right)+f\left({x}_{3}\right)+f\left({x}_{4}\right)\right)$
$=\frac{3}{2}\left(\left(49-{\frac{5}{2}}^{2}\right)+\left(49-\left(4{\right)}^{2}\right)+\left(49-{\frac{11}{2}}^{2}\right)+\left(49-\left(7{\right)}^{2}\right)$
$=\frac{3}{2}\left(42.75+33+18.75+0\right)$
$=\frac{3}{2}\left(94.5\right)$
$=141.75$
Therefore, the area under the given curve over $\left[1,7\right]$ by the right-end points approximation is 141.75.
(b) The left end-points are ${x}_{0}=1,{x}_{1}=\frac{5}{2},{x}_{2}=4,{x}_{3}=\frac{11}{2}.$
Find the area under the curve over $\left[1,7\right]$ by using the left-end points as follows,
${\int }_{1}^{7}\left(49-{x}^{2}\right)dx\approx \mathrm{\Delta }x\left(f\left({x}_{1}\right)+f\left({x}_{2}\right)+f\left({x}_{3}\right)+f\left({x}_{4}\right)\right)$
$=\frac{3}{2}\left(\left(\left(49-\left(1{\right)}^{2}\right)+49-\left(\frac{5}{2}{\right)}^{2}\right)+\left(49-\left(4{\right)}^{2}\right)+\left(49-\left(\frac{11}{2}{\right)}^{2}\right)$
$=\frac{3}{2}\left(48+42.75+33+18.75\right)$
$=\frac{3}{2}\left(142.5\right)$
$=213.75$
Therefore, the area under the given curve over $\left[1,7\right]$ by the left-end points approximation is 213.75.