(a) Given that, the function is \(f(x) = 49 - x^{2}\) on the interval \([1, 7], n = 4.\)

It is known that, the formula for the Riemann-sum is \(\int_{a}^{b}f(x)dx=\Delta x \sum_{i=0}^{\infty} f(xi),\) where \(\Delta_{x} = \frac{b-a}{n}.\)

Obtain the value of \(\Delta_{x} = \frac{b-a}{n}\)

\((a = 1, b = 7, n = 4)\)

\(\Delta_{x} = \frac{7-1}{4}\)

\(= \frac{6}{4}\)

\(= \frac{3}{2}\)

Divide the interval \([1, 7]\) into \(n = 4\) sub-intervals with length \(\Delta_{x} = \frac{3}{2}\) as \([1, \frac{5}{2}], [\frac{5}{2}, 4] [4, \frac{11}{2}], [\frac{11}{2}, 7], [\frac{11}{2}, 7]\) 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 \([1,7]\) by using the right-end points as follows,

\(\int_{1}^{7}(49-x^{2})dx \approx \Delta x(f(x_{1})+f(x_{2})+f(x_{3})+f(x_{4}))\)

\(= \frac{3}{2} ((49 - \frac{5}{2}^2) + (49 - (4)^2) + (49 - \frac{11}{2}^2) + (49 - (7)^2))\)

\(= \frac{3}{2} (42.75 + 33 + 18.75 + 0)\)

\(= \frac{3}{2} (94.5)\)

\(= 141.75\)

Therefore, the area under the given curve over \([1,7]\) 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 \([1, 7]\) by using the left-end points as follows,

\(\int_{1}^{7}(49-x^{2})dx \approx \Delta x(f(x_{1})+f(x_{2})+f(x_{3})+f(x_{4}))\)

\(= \frac{3}{2}(((49 - (1)^{2}) + 49 - (\frac{5}{2})^{2}) + (49 - (4)^2) + (49 - (\frac{11}{2})^{2}))\)

\(= \frac{3}{2} (48 + 42.75 + 33 + 18.75)\)

\(= \frac{3}{2} (142.5)\)

\(= 213.75\)

Therefore, the area under the given curve over \([1,7]\) by the left-end points approximation is 213.75.