第4章4.54.5.

1．下列积分值等于1的是()

A.eq\i\in(0,1,)xdx B．eq\i\in(0,1,)(x＋1)dx

C.eq\i\in(0,1,)1dx D．eq\i\in(0,1,)eq\f(1,2)dx

解析：eq\i\in(0,1,)1dx＝1－0＝1.

答案：C

2.eq\i\in(0,1,)(ex＋2x)dx＝()

A．1 B．e－1

C．e D．e＋1

解析：eq\i\in(0,1,)(ex＋2x)dx＝(e1＋12)－(e0＋02)

＝(e1＋1)－e0＝e.

答案：C

3.eq\i\in(0,3,)|x2－4|dx＝()

A.eq\f(21,3) B．eq\f(22,3)

C.eq\f(23,3) D．eq\f(25,3)

解析：eq\i\in(0,3,)|x2－4|dx＝eq\i\in(0,2,)(4－x2)dx＋eq\i\in(2,3,)(x2－4)dx＝eq\b\lc\[\rc\](\a\vs4\al\co1(\b\lc\(\rc\)(\a\vs4\al\co1(4×2－\f(1,3)×23))－\b\lc\(\rc\)(\a\vs4\al\co1(4×0－\f(1,3)×03))))＋eq\b\lc\[\rc\](\a\vs4\al\co1(\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,3)×33－4×3))－\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,3)×23－4×2))))＝eq\f(23,3)，故选C.

答案：C

4．若eq\i\in(－a,a,)x2dx＝18(a0)，则a＝____________.

解析：eq\i\in(－a,a,)x2dx＝eq\f(a3,3)－eq\f(?－a?3,3)＝18，即a＝3.

答案：3

5．求下列定积分：

(1)eq\i\in(1,2,)eq\f(2x2＋x＋1,x)dx；

(2)eq\i\in(0,π,)eq\r(2)sineq\b\lc\(\rc\)(\a\vs4\al\co1(x＋\f(π,4)))dx.

解：(1)eq\i\in(1,2,)eq\f(2x2＋x＋1,x)dx＝eq\i\in(1,2,)eq\b\lc\(\rc\)(\a\vs4\al\co1(2x＋\f(1,x)＋1))dx＝eq\i\in(1,2,)2xdx＋eq\i\in(1,2,)eq\f(1,x)dx＋eq\i\in(1,2,)1dx＝4－1＋ln2－ln1＋2－1＝4＋ln2.

(2)∵eq\r(2)sineq\b\lc\(\rc\)(\a\vs4\al\co1(x＋\f(π,4)))＝eq\r(2)eq\b\lc\(\rc\)(\a\vs4\al\co1(sinx·\f(\r(2),2)＋cosx·\f(\r(2),2)))＝sinx＋cosx，(－cosx＋sinx)′＝sinx＋cosx，

∴eq\i\in(0,π,)eq\r(2)sineq\b\lc\(\rc\)(\a\vs4\al\co1(x＋\f(π,4)))dx＝eq\i\in(0,π,)(sinx＋cosx)dx＝(－cosπ＋sinπ)－(－cos0＋sin0)＝2.