Câu hỏi

Chứng minh bất đẳng thức sau: 4 2 π < ∫ 0 1 1 + x 3/2 1 − x 2 d x < 6 π

Chứng minh bất đẳng thức sau: $\frac{π}{4 2} < \int_{0}^{1} \frac{1 - x ^{2}}{1 + x ^{3/2}} d x < \frac{π}{6}$

R. Robo.Ctvx9

Giáo viên

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Giải thích

Do x ∈ [ 0 , 1 ] ta có 1 + x 3/2 1 − x 2 > 1 + 1 1 − x 2 = 2 1 1 − x 2 Đặt x=sint ⇒ ∫ 0 1 1 + x 3/2 1 − x 2 d x > 2 1 ∫ 0 1 1 − x 2 d x = 2 1 ∫ 0 π /2 cos 2 t d t = 2 1 ∫ 0 π /2 2 1 + cos 2 t d t = 4 2 π Mặt khác x 3/2 ≥ x 2 ∀ x ∈ [ 0 , 1 ] ⇒ 1 + x 3/2 1 − x 2 < 1 + x 2 1 − x 2 ⇒ ∫ 0 1 1 + x 3/2 1 − x 2 d x < ∫ 0 1 1 + x 2 1 − x 2 d x Sử dụng bất đẳng thức CBS ta có Từ kết quả trên ⇒ 4 2 π < ∫ 0 1 1 + x 3/2 1 − x 2 d x < 6 π

Do $x \in [0, 1]$ ta có $\frac{1 - x ^{2}}{1 + x ^{3/2}} > \frac{1 - x ^{2}}{1 + 1} = \frac{1}{2} 1 - x^{2}$

Đặt x=sint

$\Rightarrow \int_{0}^{1} \frac{1 - x ^{2}}{1 + x ^{3/2}} d x > \frac{1}{2} \int_{0}^{1} 1 - x^{2} d x = \frac{1}{2} \int_{0}^{π /2} cos^{2} t d t = \frac{1}{2} \int_{0}^{π /2} \frac{1 + cos 2 t}{2} d t = \frac{π}{4 2}$

Mặt khác $x^{3/2} \geq x^{2} \forall x \in [0, 1]$

$\Rightarrow \frac{1 - x ^{2}}{1 + x ^{3/2}} < \frac{1 - x ^{2}}{1 + x ^{2}} \Rightarrow \int_{0}^{1} \frac{1 - x ^{2}}{1 + x ^{3/2}} d x < \int_{0}^{1} \frac{1 - x ^{2}}{1 + x ^{2}} d x$

Sử dụng bất đẳng thức CBS ta có

$open parentheses integral subscript 0 superscript 1 square root of fraction numerator 1 minus x squared over denominator 1 plus x to the power of 3 divided by 2 end exponent end fraction end root d x close parentheses squared less than open parentheses integral subscript 0 superscript 1 square root of fraction numerator 1 minus x squared over denominator 1 plus x squared end fraction end root d x close parentheses squared equals open parentheses integral subscript 0 superscript 1 square root of 1 minus x squared end root. fraction numerator 1 over denominator square root of 1 plus x squared end root end fraction d x close parentheses less or equal than integral subscript 0 superscript 1 open parentheses 1 minus x squared close parentheses d x integral subscript 0 superscript 1 fraction numerator 1 over denominator square root of 1 plus x squared end root end fraction d x equals open square brackets open parentheses x minus x cubed over 3 close parentheses a r c t g x close square brackets large vertical line subscript 0 superscript 1 equals 2 over 3 a r c t g 1 equals 2 over 3. pi over 4 equals pi over 6 rightwards double arrow integral subscript 0 superscript 1 square root of fraction numerator 1 minus x squared over denominator 1 plus x to the power of 3 divided by 2 end exponent end fraction end root d x less than square root of pi over 6 end root$

Từ kết quả trên $\Rightarrow \frac{π}{4 2} < \int_{0}^{1} \frac{1 - x ^{2}}{1 + x ^{3/2}} d x < \frac{π}{6}$