Ngô Quốc Anh

February 28, 2008

BĐT tích phân

Filed under: Các Bài Tập Nhỏ, Giải Tích 2 — Ngô Quốc Anh @ 13:40

Let [0,1]->R a differentiable function with f' continous so that :

int_{0}^{1}f(x)dx= int_{0}^{1}xf(x)dx=1.

Prove that int_{0}^{1}(f'(x))^{2}dx geq 30.

Solution. Integrating by parts we get that

intlimits_{0}^{1}xf'(x)dx=f(1)-1=alpha,

and

intlimits_{0}^{1}x^{2}f'(x)dx=f(1)-2=beta.

Thus

(beta-alpha)^{2}=1=left(intlimits_{0}^{1}(x-x^{2})f'(x)dxright)^{2}leq intlimits_{0}^{1}(x-x^{2})^{2}dxcdot intlimits_{0}^{1}f'(x)^{2}dx.

It follows that intlimits_{0}^{1}f'(x)^{2}dxgeq 30.

1 Comment »

  1. […] BĐT tích phân […]

    Pingback by Lại là BĐT tích phân « Ngô Quốc Anh — March 19, 2008 @ 0:25


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