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111 证明 令x?, 则dx??2dt, 当x?x时t?, 当x?1时t?1, 于是
txt
1111dx1x??(?)dt?dt, ?x1?x2?1?2211t1?tx1?2t1而
1?x1dx??1x所以 ?x. 21?x1?x2
?1x1121?t1dxdt??1x112dx,
7. 证明:
?1m1nnmx(1?x)dx?x(1?x)dx. 00?
11 证明 令1?x?t , 则?0xm(1?x)ndx???1(1?t)mtndt??0(1?t)mtndt??0xn(1?x)mdx, 即?0xm(1?x)ndx??0xn(1?x)mdx. 8. 证明: 证明 而 所以
?nn?0sinxdx?2?02sinxdx.
1110??nnn?0sinxdx??02sinxdx???sinxdx,
2???2?n?sin??令x???t0nnn2xdx??sin(??t)(?dt)??0sintdt??02sinxdx,
2?0sin?nxdx?2?02sinnxdx.
a?1? 9. 设f(x)是以l为周期的连续函数, 证明?a 证明 已知f(x?l)?f(x). 而 所以 因此?aa?1f(x)dx的值与a无关.
?aa?1f(x)dx??af(x)dx??0f(x)dx??l0la?lf(x)dx??0f(x)dx??lla?lf(x)dx??0f(x)dx,
a?la?laa令x?t?laf(x)dxf(t?l)dt?f(x?l)dx??0?0?0f(x)dx,
?aa?1f(x)dx??0f(x)dx.
lf(x)dx的值与a无关.
x 10. 若f(t)是连续函数且为奇函数, 证明?0f(t)dt是偶函数; 若f(t)是连续函数且为偶函数, 证明?0f(t)dt是奇函数.
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证明 设F(x)??0f(t)dt.
若f(t)是连续函数且为奇函数, 则f(?t)??f(t), 从而
?xxx令t??ux F(?x)??0f(t)dtf(?u)(?1)du?f(u)dx??0?0?0f(x)dx?F(x),
x即F(x)??0f(t)dt是偶函数.
若f(t)是连续函数且为偶函数, 则f(?t)?f(t), 从而
?xxx令t??ux F(?x)??0f(t)dtf(?u)(?1)du??f(u)dx???0?0?0f(x)dx??F(x),
x即F(x)??0f(t)dt是奇函数.
11. 计算下列定积分: (1)?0xe?xdx; 解
?x?x?x?0xedx???0xde??xee1x1110??0e?xdx??e?1?e?x110?1?2e?1.
(2)?1xlnxdx; 解
?1xlnxdx?2?e1e212lnxdx?xlnx?212e1?1e211212x?dx?e?x?20x241?(e2?1). 14e (3)?0?tsin?tdt(?为常数); 解
?02??tsin?tdt????012??tdcos?t??2?1?tcos?t2??0??cos?tdt ?0?12? ?? (4)??342??22?1?2sin?t?0??2??2.
?xsinxdx;
解
??sin2xdx????34?x???343xdcotx??xcotx4????3cotxdx???13?4??lnsinx34???34
1313 ?(?)??ln.
49224lnxdx; (5)?1x18|28
解
?14lnxxdx?2?1lnxdx?2xlnx444141?2?141x?dx x ?8ln2?2?111xdx?8ln2?4x?4(2ln2?1).
(6)?0xarctanxdx;
1112111212arctanxdx?xarctanx?x?dx ??20202021?x1?1?111?1??1 ???(1?)dx??(x?arctanx)??(1?)??.
0828201?x282442 解
?0xarctanxdx?1 (7)?02e2xcosxdx; 解
??2e2x0?cosxdx???2e2xdsinx?e2x02x??sinx20?2?02e2xsinxdx ?4?2e2x0? ?e?2?所以
?2e2xdcosx?e?0?20?2ecosx?cosxdx?e?2?4?02e2xcosxdx
???22e2x0?1cosxdx?(e??2),
5于是
(8)?1xlog2xdx; 解
?12xlog2xdx?12212logxdx?xlog2x22?122121?1221x?dx 2?1xln2 ?2??112?x2ln22?2?3. 4ln2 (9)?0(xsinx)2dx; 解
2?0(xsinx)dx??1?213x(1?cos2x)dx?x?206?0?0?1?2xdsin2x 4?0?4?0? ? ?e?31?xsin2x646?xcos2x4?031???0sin2x?2xdx4?0??316xdcos2x?0?0 .
?311??3?1??0cos2xdx???sin2x4648??3?6?4 (10)?1sin(lnx)dx; 解法一
令lnx?t1tsin(lnx)dx?1?0sint?edt.
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因为
ttt?0sint?edt??0sintde?esint11110??0etcostdt
101 ?e?sin1??0costdet?e?sin1?etcost ?e?sin1?e?cos1?1??0etsintdt, 所以 因此
1??0etsintdt
1?0ee1t1sintdt?(e?sin1?e?cos1?1).
21sin(lnx)dx?(e?sin1?e?cos1?1). ?12 解法二
?1sin(lnx)dx?x?sin(lnx)eeee1??1x?cos(lnx)?dx?e?sin1??1cos(lnx)dx 1xe ?e?sin1?x?cos(lnx)e1??1x?sin(lnx)?dx 1xe ?e?sin1?e?cos1?1??0sin(lnx)dx, 故
1sin(lnx)dx?(e?sin1?e?cos1?1). ?12eee (11)?1|lnx|dx; 解
?1e1e1|lnx|dx??1lnxdx?1lnxdx??xlnx1eee???xlnxe1??1dx??1dx
e1e111 ???e?(1?)?(e?1)?2(1?).
eee (12)?(1?x0 解
12m)2dx(m为自然数); 令x?sint?m?1dx?02costdt.
2cosn0?0(1?x12m)2 根据递推公式?m122(1?x)0?n?1?2cosn?2xdx, xdx??n0
??m?m?2?m?4 ? ? ? 5?3?1?? m为奇数?m?1m?1m?36422. dx??mm?2m?4642??? ? ? ? ?? m为偶数753?m?1m?1m?3 (13)Jm??0xsinmxdx(m为自然数). 解 因为
??令x???t0mmmxdx??(??t)sin(??t)(?1)dt??0?sintdt??0tsintdt,
??0xsin?m20|28
所以 Jm??xsinmxdx?0sinmxdx??2?02sinmxdx???02sinmxdx(用第8题结果). ?022?n?1?n22sinn?2xdx, 根据递推公式?sinxdx??0n0?m?1m?3m?5531?2?m?m?2?m?4 ? ? ? 6?4?2?2 m为偶数 Jm??.
?m?1?m?3?m?5 ? ? ? 6?4?2? m为奇数753?mm?2m?4 习题5?7
1. 判别下列各反常积分的收敛性, 如果收敛, 计算反常积分的值:
??dx (1)?; 41x 解 因为
??dx1?3??1?311 ???x?lim(?x)??, 411x???3333x所以反常积分? (2)?1??????????1??dx1dx收敛, 且?1x4?3. x4dxx;
?? 解 因为?1??dxx?2x??1?lim2x?2???, 所以反常积分?1x?????dxx发散.
(3)?0e?axdx(a>0); 解 因为
???ax1?axedx??e0a???0111?lim(?e?ax)??, x???aaa????1所以反常积分?0e?axdx收敛, 且?e?axdx?.
0a (4)?0e?ptchtdt(p>1); 解 因为
???pt1??(1?p)techtdt?[e020????11(1?p)t1?(1?p)t?e?(1?p)t]dt?[e?e]21?p1?p????0?pp2?1,
所以反常积分?0e?ptchtdt收敛, 且?0e?ptchtdt? (5)?0e?ptsin?tdt(p?0, ??0); 解
????pp?12.
???0e?ptsin?tdt??1????0e?ptdcos?t