Question

求高数大神解题

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Answer 1

(1)
∫(2/π->+∞) (1/x^2) sin(1/x) dx
= [cos(1/x)] |(2/π->+∞)
= 1 -0
=1

(2)
∫(0->1) ln(1+x)/(2+x)^2 dx
=-∫(0->1) ln(1+x) d[1/(2+x)]
=-[ln(1+x)/(2+x)]|(0->1) + ∫(0->1) dx/[(1+x)(2+x)]
=-(1/3)ln2 + ∫(0->1) [1/(x+1) - 1/(x+2)] dx
=-(1/3)ln2 + [ln|(x+1)/(x+2)|] |(0->1)
=-(1/3)ln2 + ln(4/3)

(3)
y^2 =2x (1)
y=x-4 (2)

sub (2) into (1)
y^2 = 2(y+4)
y^2-2y-4=0
y = 1+√5 or 1-√5

Area
=∫(1-√5, 1+√5) [ (y+4 )- (1/2)y^2 ] dy
=[ (1/2)y^2+4y - (1/6)y^3 ] | (1-√5, 1+√5)
=[(1/2)(1+√5)^2 +4(1+√5) - (1/6)(1+√5)^3]
-[(1/2)(1-√5)^2 +4(1-√5) - (1/6)(1-√5)^3]
=2√5 +8√5 - (1/3) (3√5 + 5√5 )
=(22/3)√5

Answer 2

按图片顺序 第一题：1 第二题：1 /6 第三题：12