2012-2013高等数学第一学期期末考试（张家港校区理工类）试卷及(2)

1?6t21?原式=?dx??dt?61?dt………………………..…………….1 2??32001?t20??1?t?1?x3?=6? ………………………..………………….1

2121?sinx??cosx?x-sinx5.解：由题意得. f(x)=?........................................ 1 ?=2xx???xf?(x)dx=?xdf(x)=xf(x)-?f(x)dx ........................................ 2

cosx?x-sinxsinx-+C ........................................ 1

x2x2sinx+C. ....................................... 1 =cosx-x=x?1dy(arctant)?1?t21???? ........................................ 2 6. 解：

tdx[ln1?t2]?t21?t1)??1(22dy1?ttt????3? ..............................................................3

tdx2[ln1?t2]?t1?t22四、证明：

?a?af(x)dx??f(x)dx??f(x)dx ..................................................... 1

?a00aaa000a令x??t,?f(x)dx??f(?t)d(?t)??f(?t)dt??f(?x)dx ..................2

?a0??f(x)dx=?f(x)?f(?x)dx ..................................................... 1

-a0aa?cosxcosxcosx4dx?(???41?ex?01?ex?1?e-x)dx ..................................................... 2

4?? ? =?40cosxdx ..................................................... 1

2 ..................................................... 1 2五、解：令x?tx0?u

xxx?tf?x?t?dt??0?x?u?f?u?du?x?0f?u?du??0uf?u?du.............................2

所以

?0f?t?dt?x?x?0f?u?du??0uf?u?du

xxx求导得：

f?x??1??0f?u?du?xf?x??xf?x??1??0f?u?du .........................2

xx第 6 页 共 7 页

求导:

f??x??f?x? ............................ ................................. ......................................1

?f?x??cex............................ ................................. .....................................1

由等式f?x??1??0f?u?du得f?0??1 则c?1 ?fx?x??xe................2

六、解：（1）A=?10y2dy................... ................................. ......................................2 2y311=................... ................................. ......................................1

606（2）V=

?120?(1-2x)2dx ................. ......................................2

1423?x2+x2)2=................. ......................................2 =?(x-3120七、证明：

由积分中值定理：????0,?，使得

??1?3?f(1)?3?e1231-x0221f(x)dx?3?e1-?f(?)(?0)?e1-?f(?) ......................................... 2

3设 F(x)?e1?xf(x) .................................................... 2 则F(x)在??,1?上连续，在??,1?内可导

且F(?)?f(1)?F(1) .................................................... 2

由罗尔定理，至少存在一点???0,??(0,1)使 F?(?)?0 .................................. 2 即f'(?)?2?f(?) .................................................... 1

2??1?3?第 7 页 共 7 页

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