Question

怎么求由曲线y=x^2-1,直线x=2,y=1所围成的平面图形面积？

Answer 1

y=x^2-1和y=1交点(√2,1)
所以S=∫(√2到2)[(x^2-1)-1]dx
=∫(√2到2)(x^2-2)dx
=(x^3/3-2x)(√2到2)
=(8/3-4)-(2√2/3-2√2)
=(4/3)√2-4/3,1,lim((1+x)^5-(1+5x))/(x^2+x^5)
=lim(x^5+5x^4+10x^3+10x^2)/(x^2+x^5)
=lim(10x^2)/(x^2)
=10
=[(1+x)^0.5-(1+sinx)^0.5]/x^3
=(x-sinx)/[(1+x)^0.5+(1-sinx)^0.5]x^3
=[(x^3)/6]/[(1+x)^0.5+(1-sinx)^0.5]x^3
=1/6[(1+x)^0.5+(1-sinx)^0.5]
=1/[6*(1+1)]
=1/12,2,