answer:{eq}begin{align} &int_1^3 (x+1)^2 e^{-x} dx &= left[(x+1)^2 int_1^3 e^{-x} , dx-int_1^3 left(frac{d}{dx}(x+1)^2 int_1^3 e^{-x} , dxright), dxright] (text{Integrate by Parts}) &= left[(x+1)^2 frac{e^{-x}}{-1} , -int_1^3 left(2(x+1), frac{e^{-x}}{-1} ,right), dxright] &=-(x+1)^2e^{-x} , +left[int_1^3 left(2(x+1), e^{-x} ,right), dxright] (text{Integrate the second term by Parts again }) &=-(x+1)^2e^{-x}, +left[(x+1) int_1^3 e^{-x} , dx-int_1^3 left(frac{d}{dx}(x+1) int_1^3 e^{-x} , dxright), dxright] &=-(x+1)^2e^{-x} , +left[(x+1) frac{e^{-x}}{-1} ,-int_1^3 frac{e^{-x}}{-1} , dxright] &=-(x+1)^2e^{-x} , +left[- (x+1)e^{-x} ,+int_1^3 e^{-x} , dxright] &=left[-(x+1)^2e^{-x} , - (x+1)e^{-x} ,- e^{-x} right]_1^3 &=-e^{-x}left[(x+1)^2 , + (x+1) ,+ 1 right]_1^3 &=e^{-x}left[(x+1)^2 , + (x+1) ,+ 1 right]_3^1 &=e^{-1}left[(1+1)^2 , + (1+1) ,+ 1 right]-e^{-3}left[(3+1)^2 , + (3+1) ,+ 1 right] , (text{ Put the limits of integration } ) &=e^{-1}left[4 , +2 ,+ 1 right]-e^{-3}left[16 , +4 ,+ 1 right] &=0.26424 - 0.10540 &=boxed{0.15884} end{align} {/eq}

answer:The expected gain from guessing on a question depends on the number of answer choices. If there are 4 answer choices, the expected gain from guessing is: E(gain) = frac{1}{4} times 4 + frac{3}{4} times (-1) = -frac{1}{4} This means that on average, you will lose 0.25 marks for each question you guess on. Therefore, it is not a good idea to randomly guess the answer to a question if you have no idea about the correct answer. However, if you are able to eliminate one or more of the answer choices, the expected gain from guessing increases. For example, if you are able to eliminate two of the answer choices, the expected gain from guessing becomes: E(gain) = frac{1}{2} times 4 + frac{1}{2} times (-1) = frac{1}{2} This means that on average, you will gain 0.5 marks for each question you guess on. Therefore, it may be a good idea to guess the answer to a question if you are able to eliminate one or more of the answer choices. Ultimately, the decision of whether or not to guess on a question is a personal one. You should consider your own risk tolerance and the potential benefits and drawbacks of guessing.

answer:We can rewrite the given series as follows: sum_{k=0}^infty -frac{57}{89}left(frac{1}{sqrt{2}}right)^k = -frac{57}{89}sum_{k=0}^infty left(frac{1}{sqrt{2}}right)^k This is a geometric series with first term a = -frac{57}{89} and common ratio r = frac{1}{sqrt{2}}. Since |r| = frac{1}{sqrt{2}} < 1, the series converges. The sum of a geometric series is given by the formula: S = frac{a}{1-r} Substituting the values of a and r, we get: S = frac{-frac{57}{89}}{1-frac{1}{sqrt{2}}} = -frac{57}{89}left(frac{sqrt{2}}{sqrt{2}-1}right) Simplifying this expression, we get: S = -frac{57}{89}left(frac{sqrt{2}}{sqrt{2}-1}right)left(frac{sqrt{2}+1}{sqrt{2}+1}right) S = -frac{57}{89}left(frac{2+sqrt{2}}{(sqrt{2})^2-1^2}right) S = -frac{57}{89}left(frac{2+sqrt{2}}{2-1}right) S = -frac{57}{89}left(2+sqrt{2}right) Therefore, the sum of the given series is -frac{57}{89}left(2+sqrt{2}right). The answer is -frac{57}{89}left(2+sqrt{2}right)

answer:The general form of the solution to this recursive equation can be derived as: p_n(x) = frac{1}{n!}a(xk)^n + bsum_{m=0}^{n-1}frac{1}{m!}(xk)^m To verify this formula, we perform induction. The base case, n=0, holds since: p_0(x) = a = frac{1}{0!}a(xk)^0 + bsum_{m=0}^{-1}frac{1}{m!}(xk)^m For the inductive step, assume the formula is true for n, then: kint_0^x p_n(t) dt + b = kint_0^x left(frac{1}{n!}a(tk)^n + bsum_{m=0}^{n-1}frac{1}{m!}(tk)^mright) dt + b After integrating and simplifying, we get: =frac{1}{(n+1)!}a(xk)^{n+1} + bsum_{m=0}^{n-1}frac{1}{(m+1)!}(xk)^{m+1} + b =frac{1}{(n+1)!}a(xk)^{n+1} + bsum_{m=1}^{n}frac{1}{m!}(xk)^{m} + b =frac{1}{(n+1)!}a(xk)^{n+1} + bsum_{m=0}^{n}frac{1}{m!}(xk)^{m} which is p_{n+1}(x), verifying the formula for all n. Notably, as n increases without bound, the polynomial sequence converges to the exponential function be^{kx}. This convergence can be understood since: kint_0^x be^{kt} dt + b = be^{kx} Thus, the polynomial sequence approaches the function be^{kx} as n tends to infinity.