第六章 一元微积分的应用 一.?选择题?1.?设?f(x)在(－￥,?+￥)内可导,?且对任意?x 1,?x2,?x1?>?x2?时,?都有?f(x1)?>?f(x2),?则?(a)?对任意x,? 0?)?(?' >?x?f? (b)?对任意?x,? 0?)?(?' ￡?x?f? (c)?函数?f(－x)单调增加? (d)?函数－f(－x)单调增加 解.?(a)?反例:? 3?)?(? x?x?f =? ,?有? 0?)?0?(?' =?f? ??(b)?显然错误.?因为? 0?)?(?' ￡?x?f? ,?函数单减??(c)?反 例:? 3?)?(? x?x?f =? ,? 3?)?(? x?x?f - = - 单调减少??排除(a),?(b),?(c)后,?(d)为答案.?具体证明如下:?令?F(x)?=?－f(－x),?x1?>?x2,?－x1?<?－x2.?所以?F(x1)?=－f(－x1)?>?－f(－x2)?=?F(x2).? 2.?设?f(x)在[－p,?+p]上连续,?当?a?为何值时, ò - - = p p? dx?nx?a?x?f?a?F? 2?]?cos?)?(?[?)?(? 的值为极小 值.? (a) ò - p p? nxdx?x?f? cos?)?(? (b) ò -p p p? nxdx?x?f? cos?)?(?1?(c) ò -p p p? nxdx?x?f? cos?)?(?2? (d) ò -p p p? nxdx?x?f? cos?)?(?2?1?解. ò - - = p p? dx?nx?a?x?f?a?F? 2?]?cos?)?(?[?)?( ò ò ò - - - + - = p p p p p p? dx?x?f?nxdx?x?f?a?nxdx?a? )?(?cos?)?(?2?cos? 2?2?2 ò ò - - + - = p p p p p? dx?x?f?nxdx?x?f?a?a? )?(?cos?)?(?2? 2?2? 为?a?的二次式.?所以当?a?= ò -p p p? nxdx?x?f? cos?)?(?1? ,?F(a)有极小值.?3.?函数?y=?f(x)具有下列特征:? f(0)?=?1??0?)?0?(?' =?f? ,?当?x1?0?时,? 0?)?(?' >?x?f? ? ? í ì > <?0?0?)?(?'?'?x?f? 0?0 > <?x?x? ,?则其图形?(a)??????????????????(b)? (c)????????????????(d)? 1?????????????????1??????????????????????1??????????????????1? 解.?(b)为答案.?4.?设三次函数? d?cx?bx?ax?x?f?y + + + = =? 2?3?)?(? ,?若两个极值点及其对应的两个极值均为相 反数,?则这个函数的图形是?(a)?关于?y轴对称? (b)?关于原点对称? (c)?关于直线?y?=?x?轴对称?(d)?以上均错 解.?假设两个极值点为?x?=?t?及?x?=?－t?(t1?0),?于是?f(t)?=－f(－t).?所以? d?ct?bt?at?d?ct?bt?at - + - = + + +? 2?3?2?3? ,?所以?b?+?d?=?0?0?2?3?)?(?'? 2 = + + =? c?bx?ax?x?f? 的根为?x?= ±t,?所以?b?=?0.?于是?d?=?0.?所以? cx?ax?x?f + =? 3?)?(?为奇函数,?原点对称.?(b)为答案.? 5.?曲线? )?2?)(?1?(? x?x?x?y - - = 与?x?轴所围图形面积可表示为 (a) ò - - -?2?0? )?2?)(?1?(? dx?x?x?x? (b) ò - -?1?0? )?2?)(?1?(? dx?x?x?x ò - - -?2?1? )?2?)(?1?(? dx?x?x?x?(c) ò - - -?1? 0? )?2?)(?1?(? dx?x?x?x ò - - +?2?1? )?2?)(?1?(? dx?x?x?x? (d) ò - -?2?0? )?2?)(?1?(? dx?x?x?x?解.? 0????????1???????2?由图知(c)为答案.? 二.?填空题?1.?函数 ò ÷ ? ? ? è ? - =?x? dt?t?x?F? 1? 1?2?)?(? (x?>?0)的单调减少区间______.? 解.? 0?1?2?)?(?' < - =? x?x?F? ,?所以?0?<?x<?4?1?.? 2.?曲线? x?x?y - =?3? 与其在? 3?1 =?x?处的切线所围成的部分被?y?轴分成两部分,?这两部分面积 之比是________.? 解.? 1?3?'? 2 - =?x?y? ,?所以切线的斜率为?k?=? 3?2?1?9?1?3 - = - × 切线方程:? 27?2?3?2 - - =? x?y? ,?曲线和切线的交点为? 3?2 - =?x? .?(解曲线和切线的联立方程得? 0?27?2?3?3 = + -?x?x? ,? 3?1 =?x?为其解,?所以可得? 0?)3?2?(?)?3?1?(? 2 = + -? x?x? ,?解得? 3?2 - =?x? .)? 比值为? 1?8?108?1?27?2?)?27?2?3?2?(? )?27?2?3?2?(3?1?0? 3?0?3?2? 3 = = + + - + + - ò ò -? dx?x?x? dx?x?x?x? 3.?二椭圆? 1?2?2?2?2 = +?b?y?a?x? ,? 1?2?2?2?2 = +?a?y?b?x? (?a?>?b>?0)之间的图形的面积______.?解.? 二椭圆的第一象限交点的x?坐标为? 2?2? b?a?ab?x + =? .?所以所求面积为 ú ? ù ê ? é - - - - = ò ò + +?2?2? 2?2?0? 0? 2?2?2?2?4? b?a?ab? b?a?ab? dx?x?b?b?a?dx?x?a?a?b?ab?s p? = ú ú ú ? ù ê ê ê ? é ÷ ? ? ? è ? - + - ÷ ? ? ? è ? - + - + +? 2?2?2?2? 0?2?2?2?0?2?2?2? 2?arcsin?2?2?arcsin?2?4? b?a?ab?b?a?ab? x?b?x?b?x?b?b?a?x?a?x?a?x?a?a?b?ab p? = ú ? ù ê ? é + - + - + + + -? )?(?2arcsin?2?)?(?2arcsin?2?4? 2?2? 2?2?2?2?2?2? 2?2?2?2? b?a?b?a?b?a?a?ab?b?a?b?a?b?a?b?ab?ab p ú ? ù ê ? é + - + - =? 2?2?2?2? arcsin?arcsin?2? b?a?a?b?a?b?ab?ab p 2?2?arcsin? b?a?a + = a 令 ú ? ù ê ? é - - - a a p p? 2?2ab?ab? =?4paba? = ÷ ? ? ? è ??b?a?abarctan?4p?4.?x? 2?+?y?2?=?a?2?绕?x=－b(b?>?a>?0)旋转所成旋转体体积_______.?解.? －b??????????????????a?由图知 ò + - - - + - =? a? dy?b?y?a?b?y?a?V? 0? 2?2?2?2?2?2? ]?)(?)[(?2p?= ò ò ò + = = -? 2? 0?2?2?0? 2?2?0? 2?2? 2?2?cos?1?8?cos?8?4?2 p p p p p? dt?tb?a?tdt?b?a?dy?y?a?b?a?=? 2?2?2? 2?4?8 p p p? ba?b?a = ×? (5)?求心脏线r=?4(1+cosq)和直线q?=?0, q?=?2 p 围成图形绕极轴旋转所成旋转体体积_____.?解.?极坐标图形绕极旋转所成旋转体体积公式 ò = ba q q q r? d?V? sin?)?(?3?所以 ò ò + = =? 2?0? 3?3? sin?)?cos?1?(?643?2?sin?)?(?3?2 p ba q q q p q q q r p? d?d?V? = p p p p q p? 160?16?3?32?3?32?0?2?)?cos?1?(4?1?3?128? 4 = × + - = + × - 三.?计算题?1.?在直线x－y?+?1=0?与抛物线? 5?4?2 + - =? x?x?y? 的交点上引抛物线的法线,?试求由两法线及 连接两交点的弦所围成的三角形的面积.?解.?由联立方程 ? í ì + - = = + -? 5?4?0?1? 2? x?x?y?y?x? 解得交点坐标? )?2?,?1?(?)?,?(? 3?3 =?y?x? ,? )?5?,?4?(?)?,?(? 2?2 =?y?x?由? 4?2?' - =?x?y? 求得二条法线的斜率分别为? 2?1?1 = =?x?k? ,? 4?1?4 - = =?x?k? .?相应的法线为?)?1(2?1?2 - = -? x?y? ,? )?4(4?1?5 - - = -? x?y? .?解得法线的交点为? )2?9,?6?(?)?,?(? 1?1 =?y?x? .? 已知三点求面积公式为? 3?2?3?2? 3?1?3?1?2?1? y?y?x?x? y?y?x?x?S - - - - ± = 所以? 4?15?3?3?2?5?5?2?1?2?1? 3?2?3?2? 3?1?3?1 = ± = - - - - ± =? y?y?x?x? y?y?x?x?S? .?2.求通过点(1,?1)的直线y?=?f(x)中,?使得 ò -?2? 0? 2?2? )]?(?[? dx?x?f?x? 为最小的直线方程. 解.?过点(1,?1)的直线为?y?=kx+?1－k?所以? F(k)?= ò - - -?2?0? 2?2? )]?1?(?[? dx?k?kx?x?= ò - + - + - + + -?2? 0? 2?2?2?3?4? ]?)?1?(?)?1?(?2?)?22?(?2?[? dx?k?x?k?k?x?k?k?kx?x?=? 2? 0?2?2?3? 2?4?5? )?1?(?)?1?(?3?2?2?4?2?5 ú ? ù ê ? é - + - + - + + -? x?k?x?k?k?x?k?k?x?k?x? =? 2?2? )?1?(?2)?1?(?4?)?22?(3?8?8?5?32? k?k?k?k?k?k - + - + - + + -? 0?3?8?3?4?)?1?(?4)?8?4?(?)?22?(3?8?8?)?(?' = - = - - - + + + - =? k?k?k?k?k?F? k?=?2?所求直线方程为? y?=?2x－1? 3.?求函数 ò - - =?2?0? )?2?(?)?(? x? t?dt?e?t?x?f? 的最大值与最小值.?解.? 0?)?2?(?2?)?(?'? 2?2?2 = - = -x?e?x?x?x?f? ,?解得? x?=?0,? x?=?2 ±?0?)?0?( =?f? ,? 2?2? 0? 1?)?2?(?)?2?( - - + = - = ± ò e?dt?e?t?f? t? , ò +￥ - - = ±￥? 0? )?2?(?)?(? dt?e?t?f? t? =1?所以,?最大值? 2?1?)?2?( - + = ±? e?f? ,?最小值? 0?)?0?( =?f? .? 4.?已知圆(x－b)?2?+?y?2?=?a?2?,?其中?b?>a?>?0,?求此圆绕?y轴旋转所构成的旋转体体积和表面积.?解.?体积 ò ò - + - + = - - - ′ =? 2?2? 2?2?2?2? cos?)?sin?(?4?sin?)?(?2?2 p p p p? tdt?a?t?a?b?t?a?b?x?dx?bx?a?x?V? a?b?a?b? 令?=? 2?2?2?2?0? 2?2? 2?4?8?cos?8? ba?ba?dt?t?ba p p p p p = × = ò 表面积:?y=?f(x)绕x?轴旋转所得旋转体的表面积为? S= ò +?b?a? dx?x?f?x?f? )?(?'?1?)?(?2? 2 p(x－b)?2?+?y?2?=?a?2?绕?y轴旋转相当于(y－b)?2?+?x?2?=?a?2?绕x?轴旋转.?该曲线应分成二枝:? 2?2? x?a?b?y - ± = 所以旋转体的表面积 ò ò - - - - - + - - + =? a?a?a?a? dx?x?a?a?x?a?b?dx?x?a?a?x?a?b?S? 2?2?2?2?2?2?2?2? )?(?2?)?(?2 p p? =? ab?dt?ab?x?a?dx?ab?a?a? 2?2?0?2?2? 4?8?4 p p p p = = - ò ò -? .?5、 设有一薄板其边缘为一抛物线(如图),?铅直沉入水中,? i.?若顶点恰在水平面上,?试求薄板所受的静压力.?将薄板下沉多深压力加倍??ii.?若将薄板倒置使弦恰在水平面上,?试求薄板所受的静压力.?将薄板下沉多深压力加倍?? 解.?i.?由图知抛物线方程为? 5?3?x?y =? .?于是 dx?x?dx?y?x?dp? 5?6?2? 2?3 = × × =? 1920?0?20?5?2?5?6?5?6? 2?5?20?0? 2?3 = × × = = ò x?dx?x?p? 假设将薄板沉到水中,?深为?h处,?此时薄板的曲线方程为?5?3? h?x?y - =? ,? dx?h?x?x?dx?y?x?dp? 5?6?2 - = × × = 由题设知? 1920?2?5?6?20 ′ = - ò +?h? h? dx?h?x?x? ,?即? 640?5?20 = - ò +?h?h? dx?h?x?x?640?)?( 3?2?)?(5?2?5?1? 20?2?3?2?5 = ú ? ù ê ? é - + - +?h?h?h?x?h?x?4?3 =?h? ,? h?=?12? ii.?由图知抛物线方程为? 5?20?3? x?y - =? .?于是? dx?x?x?dx?y?x?dp? 5?20?6?2 - = × × =? 1280?0?20?)?20?(3?2?5?120?0?20?)?20?(?5?2?5?6?20?5?6? 2?3?2?5?20?0 = - × - - × × = - = ò? x?x?dx?x?x?p? 假设将薄板沉到水中,?深为?h处,?此时薄板的曲线方程为?5?20?3? x?h?y - + =? ,? dx?x?h?x?dx?y?x?dp? 5?20?6?2 - + = × × = 由题设知? 1280?2?5?20?6?20 ′ = - + ò +?h? h? dx?x?h?x? ,?即? 2560?)?20?(?5?6?20? 2?1 = - + ò +?h?h? dx?x?h?x? h?h?x?h? 20?)?20?(5?2?5?6? 2?5 + - + ×? h?h?x?h?h? 20?)?20?)(?20?(3?2?5?6? 2?3 + - + + × - －12?+?20?+?h?=?16,???h?=?18? 四.?证明题?1.?设?f(x)为连续正值函数,?证明当?x3?0?时函数 ò ò =? x?x? dt?t?f? dt?t?tf?x? 0?0? )?()?()?( f 单调增加.? 证明.? 0?)?(? )?(?)?(?)?(?)?(? )?(?)?(?)?(?)?(?'? 2?0?0?2?0? 0?0 > ú ? ù ê ? é - = ú ? ù ê ? é ú ? ù ê ? é - = ò ò ò ò ò? x?x?x? x?x? dt?t?f? dt?t?f?t?x?x?f?dt?t?f? dt?t?tf?dt?t?f?x?x?f?x f 上述不等式成立是因为?f(x)?>?0,? t<?x.?2.?设?f(x)在[a,?b]上连续,?在(a,?b)内? 0?)?(?'?' >?x?f? ,?证明? a?x?a?f?x?f?x - - =? )?(?)?(?)?( f 在(a,?b)内单增.?证明.?假设?a<?x 1?<x2?<?b,? )?(?'?)?(?)?(?)?(? 1?1?1?1 x f? f?a?x?a?f?x?f?x = - - =? (a?< x1?<x1?)? a?x? a?f?x?f?x?f?x?f?a?x?a?f?x?f?x - - + - = - - =? 2? 1?1?2?2?2?2? )?(?)?(?)?(?)?(?)?(?)?(?)?( f? a?x? a?x?f?x?x?f - - + - =? 2? 1?1?1?2?2? )?)(?(?'?)?)(?(?' x x? )?(?)?(?'?)?)(?(?'? 1?1?2? 1?1?2?1? x?f?a?x? a?x?x?x?f f x x = = - - + - > 不等式成立是因为x 1?<x1?<x2.? 0?)?(?'?' >?x?f? 说明? )?(?'?x?f?单增,?于是? )?(?)?(?'? 1?2 x x? f?f >? .?3.?设?f(x)在[a,?b]上连续,?在(a,?b)内可导且? 0?)?(?' ￡?x?f? ,?求证: ò - =? x?a? dt?t?f?a?x?x?F? )?(1?)?(?在(a,?b)内也? 0?)?(?' ￡?x?F? .? 证明:?方法?1:?因为? 0?)?(?' ￡?x?f? ,?所以?f(x)单减.?)?(?1?)?(? )?(? 1?)?(?'? 2? x?f?a?x?dt?t?f?a?x?x?F? x?a - + - - = ò? = ò - -? x?a? dt?t?f?a?x? )?(?)?(? 1?2? + ò -? x?a? dt?x?f?a?x? )?(?)?(?1?2?= ò < - -? x?a? dt?t?f?x?f?a?x? 0?)]?(?)?(?[?)?(?1?2?所以?F(x)单减.?方法?2:?假设?a<?x 1?<x2?<?b? )?(?)?(?1?)?(? 1?1?1? 1 xf?dt?t?f?a?x?x?F? x?a = - = ò? ,??(a?< x1?<?x1)(因为?f(x)单减, x1?在区间(a,?x1)内部). ú ? ù ê ? é + - = - = ò ò ò? 21?1?2? )?(?)?(?1?)?(?1?)?(? 2?2?2? x?x?x?a?x?a dt?t?f?dt?t?f?a?x?dt?t?f?a?x?x?F? =? )]?)(?(?)?)(?(?[?1? 1?2?2?1?1?2? x?x?f?a?x?f?a?x - + - - x x <? )?(?)?(?)]?)(?(?)?)(?(?[?1? 1?1?1?2?1?1?1?2? x?F?f?x?x?f?a?x?f?a?x = = - + - - x x x 因为?f(x)单减,x2?在区间(x1,?x2)内部,?即?a?< x1?<?x1?< x2?<x2,?且? )?(?)?(? 1?2 x x? f?f <? .?4.证明方程? x?x - =1?tan? 在(0,?1)内有唯一实根.? 证明.?令? x?x?x?F + - =? 1?tan?)?(? .?F(0)?=－1?<0,?F(1)?=?tan1?>0,?所以在(0,?1)中存在x,?使?F(x)?=?0. 又因为? 0?1?cos?1?)?(?'? 2 > + =? x?x?F? (0?<?x?<?1),?所以?F(x)单增,?所以实根唯一.?5.?设?a 1,?a2,?…,?an?为?n个实数,?并满足? 0?1?2?)?1(?3? 1?2?1 = - - + + - -? n?a?aa? n?n?L? .?证明:?方程?0?)?1?2?cos(?3?cos?cos? 2?1 = - + +? x?n?a?x?a?x?a? n?L 在(0,?2 p?)内至少有一实根.? 证明.?令 =?)?(x?F? 1?2? )?12?sin(?3?3?sin?sin? 2?1 - - + +? n? x?n?a?x?ax?a? n?L 则?F(0)?=?0, = ÷ ? ? ? è ??2 p?F? 0?1?2?)?1(?3? 1?2?1 = - - + + - -? n?a?aa? n?n?L? .?所以由罗尔定理存在x?(0?< x?<?2 p? ),?使? 0?)?(?' = x?F? .?即? 0?)?1?2?cos(?3?cos?cos? 2?1 = - + + x x x? n?a?a?a? n?L 五.?作图? 1.? 1?2?2?2 - + - =? x?x?x?y? 2.? 2?)?1?(? 2? x?e?x?y - + = 解.? 1.???????????????????????????2.