已知a>0,当∫(cosx-sinx)dx(上限a,下限O)取最大值时,a的最小值为___________(要详细过程,
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已知a>0,当∫(cosx-sinx)dx(上限a,下限O)取最大值时,a的最小值为___________(要详细过程,)
![已知a>0,当∫(cosx-sinx)dx(上限a,下限O)取最大值时,a的最小值为___________(要详细过程,](/uploads/image/z/14871870-54-0.jpg?t=%E5%B7%B2%E7%9F%A5a%3E0%2C%E5%BD%93%E2%88%AB%EF%BC%88cosx-sinx%29dx%EF%BC%88%E4%B8%8A%E9%99%90a%2C%E4%B8%8B%E9%99%90O%EF%BC%89%E5%8F%96%E6%9C%80%E5%A4%A7%E5%80%BC%E6%97%B6%2Ca%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E4%B8%BA___________%28%E8%A6%81%E8%AF%A6%E7%BB%86%E8%BF%87%E7%A8%8B%2C)
∫[0->a) (cosx-sinx) dx
= cosx+sinx(0->a)
= cosa+sina-cos0-sin0
= cosa+sina-1
设f(a) = cosa+sina-1
= √2*sin(a+π/4) - 1
当sin(a+π/4) = 1
a+π/4 = π/2
a = π/4时,f(a)取得最大值
并且最大值为f(π/4) = √2*sin(π/4+π/4) - 1
= √2 - 1
∴a的值为π/4
再问: ∫[0->a) (cosx-sinx) dx = cosx+sinx(0->a) = cosa+sina-cos0-sin0 = cosa+sina-1 这个过程我不大理解,请解释一下,谢谢!
再答: cosx的积分是sinx sinx的积分是-cosx 所以积分∫(cosx-sinx) = ∫cosx dx - ∫sinx dx = sinx - (-cosx) = sinx + cosx 求得原函数后再代入上限和下限,即∫(a->b) f(x) dx = F(b) - F(a),F(x)是f(x)的原函数 = (sinx + cosx) |(上限a,下限0) = (sina + cosa) - (sin0 + cos0),上限a代入的原函数再减去下限0代入的原函数 = (sina + cosa) - (0 + 1) = sina + cosa - 1
= cosx+sinx(0->a)
= cosa+sina-cos0-sin0
= cosa+sina-1
设f(a) = cosa+sina-1
= √2*sin(a+π/4) - 1
当sin(a+π/4) = 1
a+π/4 = π/2
a = π/4时,f(a)取得最大值
并且最大值为f(π/4) = √2*sin(π/4+π/4) - 1
= √2 - 1
∴a的值为π/4
再问: ∫[0->a) (cosx-sinx) dx = cosx+sinx(0->a) = cosa+sina-cos0-sin0 = cosa+sina-1 这个过程我不大理解,请解释一下,谢谢!
再答: cosx的积分是sinx sinx的积分是-cosx 所以积分∫(cosx-sinx) = ∫cosx dx - ∫sinx dx = sinx - (-cosx) = sinx + cosx 求得原函数后再代入上限和下限,即∫(a->b) f(x) dx = F(b) - F(a),F(x)是f(x)的原函数 = (sinx + cosx) |(上限a,下限0) = (sina + cosa) - (sin0 + cos0),上限a代入的原函数再减去下限0代入的原函数 = (sina + cosa) - (0 + 1) = sina + cosa - 1
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