Consider the following. f(x)=49−x2 from x=1 to x=7,4 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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Consider the following. f(x)=49−x2 from x=1 to x=7,4 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.

mhalmantus · Accepted Answer

(a) Given that, the function is f(x)=49−x2 on the interval [1,7],n=4. It is known that, the formula for the Riemann-sum is ∫abf(x)dx=Δx∑i=0∞f(xi), where Δx=b−an. Obtain the value of Δx=b−an (a=1,b=7,n=4) Δx=7−14 =64 =32 Divide the interval [1,7] into n=4 sub-intervals with length Δx=32 as [1,52],[52,4][4,112],[112,7],[112,7] and the right end-points are x1=52,x2=4,x3=112,x4=7. Find the area under the curve over [1,7] by using the right-end points as follows, ∫17(49−x2)dx≈Δx(f(x1)+f(x2)+f(x3)+f(x4)) =32((49−522)+(49−(4)2)+(49−1122)+(49−(7)2) =32(42.75+33+18.75+0) =32(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 x0=1,x1=52,x2=4,x3=112. Find the area under the curve over [1,7] by using the left-end points as follows, ∫17(49−x2)dx≈Δx(f(x1)+f(x2)+f(x3)+f(x4)) =32(((49−(1)2)+49−(52)2)+(49−(4)2)+(49−(112)2) =32(48+42.75+33+18.75) =32(142.5) =213.75 Therefore, the area under the given curve over [1,7] by the left-end points approximation is 213.75.

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

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