大学数学题目,请数学高手作答,非常感激. 1、求导数: 函数,不定积分
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(1):y = x²sin2x
y' = sin2x • 2x + x² • 2cos2x
= 2xsin2x + 2x²cos2x
= 2x(sin2x + xcos2x)
(2):∫ xlnx dx = ∫ lnx d(x²/2)
= (x²/2)lnx - (1/2)∫ x² d(lnx)
= (1/2)x²lnx - (1/2)∫ x² • 1/x dx
= (1/2)x²lnx - (1/2)∫ x dx
= (1/2)x²lnx - (1/2)(x²/2) + C
= (x²/4)(2lnx - 1) + C
(3):∫ (2 - 3x)¹² dx
令u = 2 - 3x,du = (2)'dx - (3x)'dx = -3dx
原式= ∫ u¹² • du/(-3)
= (-1/3) • u¹³/13 + C
= (-1/39)(2 - 3x)¹³ + C
(4):
f(x) = 3x³ - 9x + 7
f'(x) = 9x² - 9
f''(x) = 18x
f'(x) = 0 => x = ±1
f''(-1) < 0,取得极大值,f''(1) > 0,取得极小值
极大值f(-1) = 13,极小值f(1) = 1
在端点,f(-3) = -47,f(3) = 61
故在x∈[-3,3]中,最大值为61,最小值为-47
y' = sin2x • 2x + x² • 2cos2x
= 2xsin2x + 2x²cos2x
= 2x(sin2x + xcos2x)
(2):∫ xlnx dx = ∫ lnx d(x²/2)
= (x²/2)lnx - (1/2)∫ x² d(lnx)
= (1/2)x²lnx - (1/2)∫ x² • 1/x dx
= (1/2)x²lnx - (1/2)∫ x dx
= (1/2)x²lnx - (1/2)(x²/2) + C
= (x²/4)(2lnx - 1) + C
(3):∫ (2 - 3x)¹² dx
令u = 2 - 3x,du = (2)'dx - (3x)'dx = -3dx
原式= ∫ u¹² • du/(-3)
= (-1/3) • u¹³/13 + C
= (-1/39)(2 - 3x)¹³ + C
(4):
f(x) = 3x³ - 9x + 7
f'(x) = 9x² - 9
f''(x) = 18x
f'(x) = 0 => x = ±1
f''(-1) < 0,取得极大值,f''(1) > 0,取得极小值
极大值f(-1) = 13,极小值f(1) = 1
在端点,f(-3) = -47,f(3) = 61
故在x∈[-3,3]中,最大值为61,最小值为-47
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非常感谢你的帮忙,请再帮忙做几道题,实太太感谢你了!
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(1):y = cosx,y' = - sinx,y'' = - cosx,y''' = sinx
(2):∫ (2^x + x²) = ∫ 2^x dx + ∫ x² dx = (2^x)/ln2 + x³/3 + C
(3):y = ax^(b + c)
y' = a(b + c)x^(b + c - 1)
(4):
lim(x→2) (x² - x + k)/(x - 2) = 3,分子能够因式分解,并且能约去x - 2项
所以(x² - x + k) = (x - 2)(x + m)
x² - x + k = x² + (m - 2)x - 2m
{ m - 2 = -1 => m = 1
{ k = - 2m => k = (-2)(1) = -2
所以k = -2
(5):∫ dx/(3 + 2x) = (1/2)∫ d(2x)/(3 + 2x) = (1/2)∫ d(3 + 2x)/(3 + 2x) = (1/2)ln|3 + 2x| + C
(6):d(?) = (2^x)ln2 dx
d(?)/dx = (2^x)ln2
?= (ln2)∫ 2^x dx = (ln2) • (2^x)/ln2 + C = 2^x + C
故d(2^x + C) = (2^x)ln2 dx,C为任意的常数
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1、求导数
导数=2x sin2x+2x^2 con2x
2.不定积分=(1/2) (ln x)^2 + c
3.t=3x-2,
(2-3x )^12= )(3x-2)^12
dx=dt/3 ,
不定积分=(1/39)(3x-2)^13+c
4
F=3X [X+(3)^0.5] [X-(3)^0.5] +7
x=3. Fmax=54+7=61,
x=-3, Fmin= -54+7=-47
导数=2x sin2x+2x^2 con2x
2.不定积分=(1/2) (ln x)^2 + c
3.t=3x-2,
(2-3x )^12= )(3x-2)^12
dx=dt/3 ,
不定积分=(1/39)(3x-2)^13+c
4
F=3X [X+(3)^0.5] [X-(3)^0.5] +7
x=3. Fmax=54+7=61,
x=-3, Fmin= -54+7=-47
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实在太感谢你了,我还有几道题能再麻烦你吗?谢谢
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