已知函数 f ( x ) = ∫ 1 x 1 + t 4 d t 已知函数f(x)=\int_{1}^{x}\sqrt{1+t^4}\,{\rm d}t 已知函数f(x)=∫1x1+t4 dt\ 则 ∫ 0 1 x 2 f ( x ) d x = 则 \int_{0}^{1}x^2f(x)\,{\rm d}x= 则∫01x2f(x)dx=
∫ 0 1 x 2 f ( x ) d x \int_{0}^{1}x^2f(x)\,{\rm d}x ∫01x2f(x)dx
= 1 3 ∫ 0 1 f ( x ) d x 3 =\frac{1}{3}\int_{0}^{1}f(x)\,{\rm d}x^3 =31∫01f(x)dx3
= 1 3 f ( x ) x 3 ∣ 0 1 − 1 3 ∫ 0 1 x 3 d f ( x ) =\frac{1}{3}f(x)x^3|_{0}^{1} -\frac{1}{3}\int_{0}^{1}x^3df(x) =31f(x)x3∣01−31∫01x3df(x)
= 0 − 1 3 ∫ 0 1 x 3 f ′ ( x ) d x =0-\frac{1}{3}\int_{0}^{1}x^3f'(x)dx =0−31∫01x3f′(x)dx
= − 1 3 × 1 4 ∫ 0 1 f ′ ( x ) d ( x 4 ) =-\frac{1}{3}\times \frac{1}{4}\int_{0}^{1}f'(x)d(x^4) =−31×41∫01f′(x)d(x4)
= − 1 12 ∫ 0 1 x 4 + 1 d ( x 4 + 1 ) =-\frac{1}{12}\int_{0}^{1}\sqrt{x^4+1}d(x^4+1) =−121∫01x4+1 d(x4+1)
= − 1 12 × 2 3 ( x 4 + 1 ) 3 2 ∣ 0 1 =-\frac{1}{12}\times\frac{2}{3}(x^4+1)^{\frac{3}{2}}|_{0}^{1} =−121×32(x4+1)23∣01
= − 1 18 ( 2 2 − 1 ) =-\frac{1}{18}(2\sqrt{2}-1) =−181(22 −1)