已知∫(上限x下限0)tf(2x-t)dt=0.5arctanx^2 ,f(1)=1 ,求∫(上限2下限1)f(x)dx
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已知∫(上限x下限0)tf(2x-t)dt=0.5arctanx^2 ,f(1)=1 ,求∫(上限2下限1)f(x)dx
已知∫(上限x下限0)tf(2x-t)dt=0.5arctanx^2 ,f(1)=1 ,求∫(上限2下限1)f(x)dx
已知∫(上限x下限0)tf(2x-t)dt=0.5arctanx^2 ,f(1)=1 ,求∫(上限2下限1)f(x)dx
结果得3/4
计算过程如下:
(1):令2x-t=u t:0->x 则u:2x->x 且dt=-du
∫(上限x下限0)tf(2x-t)dt=∫(上限x下限2x)(u-2x)f(u)dtu=∫(上限x下限0)(u-2x)f(u)dtu-∫(上限2x下限0)(u-2x)f(u)dtu=0.5arctanx^2
(2):两边求导转化一下形式就可以得到∫(上限2x下限x)f(x)={1/(1+x^4)+xf(x)}*(1/2)
带入x=1即可得出结果
答案是3/4
∫[0,x] t * f(2x-t) dt = 0.5 * arctan(x²)
代换:y = 2x-t,dy = -dt
t = 0,y = 2x;t = x,y = x
∫[2x,x] (2x-y) * f(y) (-dy) = 0.5 * arctan(x²)
(2x)∫[x,2x] f(y) dy - ∫[x,2x]...
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答案是3/4
∫[0,x] t * f(2x-t) dt = 0.5 * arctan(x²)
代换:y = 2x-t,dy = -dt
t = 0,y = 2x;t = x,y = x
∫[2x,x] (2x-y) * f(y) (-dy) = 0.5 * arctan(x²)
(2x)∫[x,2x] f(y) dy - ∫[x,2x] yf(y) dy = 0.5 * arctan(x²),下一步两边求导
d/dx ∫[x,2x] f(y) dy = (2x)'f(2x) - (x)'f(x) = 2f(2x) - f(x)
d/dx (2x)∫[x,2x] f(y) dy = 2∫[x,2x] f(y) dy + 4xf(2x) - 2xf(x)
d/dx ∫[x,2x] yf(y) dy = (2x)'(2x)f(2x) - (x)'xf(x) = 4xf(2x) - xf(x)
原本的方程变为:
{2∫[x,2x] f(y) dy + 4xf(2x) - 2xf(x)} - {4xf(2x) - xf(x)} = 0.5 * 1/(1+x⁴) * 2x
2∫[x,2x] f(y) dy - xf(x) = x/(1+x⁴)
∫[x,2x] f(y) dy = x/[2(1+x⁴)] + (1/2)xf(x),下一步将x=1代入,得
∫[1,2] f(x) dx = 1/4 + (1/2)(1)f(1) = 1/4 + (1/2) = 3/4
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