两道高等数学微积分题？

Question

如题所述

Answer 1

1、y=∫(0,x) 4^t*(1+2tln2)dt
=∫(0,x) (1/ln4+t)d(4^t)
=4^t*(1/ln4+t)|(0,x)-∫(0,x) 4^t*d(1/ln4+t)
=4^x*(1/ln4+x)-1/ln4-(1/ln4)*4^t|(0,x)
=4^x*(1/ln4+x)-1/ln4-(1/ln4)*4^x+1/ln4
=x*4^x
所以y=x*4^x与y=4x的交点为(0,0)和(1,4)
两条曲线围成的面积=∫(0,1) (4x-x*4^x)dx
=∫(0,1) 4xdx-∫(0,1) x*4^xdx
=2x^2|(0,1)-∫(0,1) x/ln4*d(4^x)
=2-(x/ln4)*4^x|(0,1)+(1/ln4)*∫(0,1) 4^xdx
=2-2/ln2+(1/ln4)^2*4^x|(0,1)
=2-2/ln2+3/(ln4)^2
2、当x≠0时，f(x)=∫(0,x) [x(x-t)f(t)+1]/xdt
=∫(0,x) (x-t)f(t)dt+∫(0,x) (1/x)dt
=x∫(0,x) f(t)dt-∫(0,x) tf(t)dt+(t/x)|(0,x)
=x∫(0,x) f(t)dt-∫(0,x) tf(t)dt+1
根据题意，f(0)=lim(x->0)f(x)=1
f'(x)=∫(0,x) f(t)dt+xf(x)-xf(x)=∫(0,x) f(t)dt
根据题意，f'(0)=lim(x->0)f'(x)=0
f''(x)=f(x)
特征方程r^2-1=0，r1=1，r2=-1
所以f(x)=C1*e^x+C2*e^(-x)
f'(x)=C1*e^x-C2*e^(-x)
将x=0代入，f(0)=C1+C2=1，f'(0)=C1-C2=0
得：C1=C2=1/2
f(x)=(1/2)*[e^x+e^(-x)]