Question

几道高等数学的题目,求大家帮忙做一下 拜托

要有过程啊,谢谢大家了
把这三题做出来也可以的 其它的几题我做了





Answer 1

1.lim(x→∞) [(x+3)/(x+6)]^[(x-1)/2]
=lim(x→∞)e^[((x-1)/2)ln ((x+3)/(x+6))]
=e^[1/2•lim(x→∞)(x-1)ln ((x+3)/(x+6))]
又∵lim(x→∞)(x-1)ln ((x+3)/(x+6))
=lim(x→∞)ln ((x+3)/(x+6)) / [1/(x-1)]
【为0/0型（lim(x→∞)ln ((x+3)/(x+6))=lim(x→∞)ln(1- 3/(x+6))=ln1=0）。用洛必达，上下同时求导】
=lim(x→∞) ((x+6)/(x+3))•((x+3)/(x+6))′ / [1/(x-1)] ′
【PS:注意（ln ((x+3)/(x+6))）′复合函数求导 】
=lim(x→∞)1/((x+3)(x+6)) / [-1/(x-1)^2]
=lim(x→∞)(x-1)^2/((x+3)(x+6))
=lim(x→∞)(x^2-2x+1)/(x^2+9x+18)【有理分式看最高此项前系数比.分母含有最高此项x^2前系数为1.分1含有最高此项x^2前系数为1.系数比为1/1=1】
=1
∴原式=e^(1/2 •1)=e^1/2

2.∫(0→1)(x^2+x-2)e^xdx
=∫(0→1)x²dx+∫(0→1)xdx-2∫(0→1)e^xdx
=[（x²e^x）|(0→1) -∫(0→1)e^xdx²] +∫(0→1)xdx-2∫(0→1)e^xdx
=[（x²e^x）|(0→1) -2∫(0→1)xe^xdx] +∫(0→1)xdx-2e^x|(0→1)
=（x²e^x）|(0→1) -∫(0→1)xe^xdx -2e^x|(0→1)
=（x²e^x）|(0→1) -∫(0→1)xde^x -2e^x|(0→1)
=（x²e^x）|(0→1) -[xe^x |(0→1)-∫(0→1)e^xdx]-2e^x|(0→1)
=（x²e^x）|(0→1) -xe^x |(0→1)+e^x|(0→1)-2e^x|(0→1)
=（x²e^x）|(0→1) -xe^x |(0→1)-e^x|(0→1)
=2e-2

3.依题意有dA=x²dx
A=∫（a→(a+1)）x²dx
=x³/3|（a→(a+1)）
=((a+1)³-a³)/3
=((a+1)-a)((a+1)²+a(a+1)+a²)/3
=(3a²+3a+1)/3
=a²+a+1/3
=(a+1/2)²+1/12
当a=-1/2时。Amin=1/12
此时dV=∏(x²)²dx=∏x^4dx
V=∫(-1/2→1/2)∏x^4dx
=∏/5 • x^5|(-1/2→1/2)
=∏/80

Answer 2

问答题第一题用微积分的倒数做，不会在电脑上打符号，所以请见谅