7.6, #8. LEFT= 0.815997, RIGHT= 0.753497, and TRAP= 0.785724. The actual value of the integral is π/4. Here LEFT is an over estimate and RIGHT is an under estimate since the integrand 1/(1+x^{2}) is decreasing on [0, 1]. Thus 0.753497 < π/4 < 0.815997 and 3.013988 < π < 3.263988.

7.6, #10. The function g is decreasing and concave-up on the interval [0, 1] of integration. So RIGHT<TRAP<LEFT since TRAP is the average of LEFT and RIGHT. We also have MID<TRAP since g is concave-up, and RIGHT<MID since g is decreasing. So LEFT= 0.664, TRAP= 0.633, MID= 0.632, and RIGHT= 0.601. The true value lies between TRAP and MID, that is, 0.633 > ∫_{0}^{1} g(t) dt > 0.632

7.6, #21. It may help to think about this problem first in the case where 100000 is replaced by 4.

∑_{k=1}^{100000}(1/k) is a left sum for ∫_{1}^{100001}(1/x) dx with N = 100000. ∑_{k=2}^{100000}(1/k) is a right

sum for ∫_{1}^{100000}(1/x) dx with N = 999999. The integrand is decreasing, so

\begin{aligned}∑_{k=1}^{100000}(1/k) & ≥ ∫_{1}^{100001} \frac{dx}{x} = ln 100001 = 11.51293546 \\ ∑_{k=2}^{100000}(1/k) & ≤ ∫_{1}^{100000} \frac{dx}{x} = ln 100000 = 11.51292546 \end{aligned}

Thus 11.512 ≤ ∑_{k=1}^{100000}(1/k) ≤ 1 + 11.513 (the term 1 comes from 1/k for k = 1). The nearest integer is 12.

7.7, #2. Simpson’s rule with n = 10 gives the approximation 0.6931473747 for ∫_{1}^{2}(1/x) dx. On the other hand, the integral is ln 2 = 0.6931471806. The approximation is larger than the integral with the error being 0.0000001941.

7.7, #7. The approximations are 1.089429413 for n = 10, 1.089429413 for n = 50, 1.089429413 for n = 250. The approximations are the same to 9 decimal places, so we take 1.089 as the value of ∫_{0}^{1} \sqrt{1 + x^{4}} dx.

7.7, #8. The approximations are 29.09408256 for n = 10, 29.09346052 for n = 50, 29.09346052 for n = 250. The last two approximations are the same to 8 decimal places, so we take 29.093 as the value of ∫_{0}^{10} ln(z^{2} + 1) dz.