麻煩幫我解以下微積分題目~~麻煩了!!謝謝

Question

1.f(x)=(x-3)2,[2,5](a)Find the average value of f on the given interval.(b)Find c such that fave=f(c)2.試求兩曲線y=sinx,y=cosx與兩條直線x=0,x=pi/2所圍成區域之面積3.Let f(x)=3/64x(16-x)1/2 for 0<=x<=4 and f(x)=0 for all other values of x.(a)Verify that f is a probability density function.(b)Find P(X<2)4.Let f(x)=5tanx. Find the value of f\'(pi/4).5.Evaluate the integral ∫cos x1/2dx

Answer 1

1.(a)平均值=(∫25 f(x) dx )/(5-2)= (∫25 (x-3)² dx)/3= ([(1/3)(x-3)³]25 )/3= ((8/3)-(-1/3))/3=3/3= 1(b) (c-3)² =1c = 2 or 42.面積=∫0π/4 (cosx-sinx) dx +∫π/4π/2 (sinx-cosx) dx= [sinx+cosx]0π/4 + [- cosx- sinx]π/4π/2 = (√2)-(0+1) + (0-1)-(-√2)= 2√2 - 23. f(x)=3/64x(16-x)1/2有點怪怪的，可能少了一個 ² 不然積出來不是1。(a) f(x)=(3/64)x(16-x² )1/2 , 0≦x≦4 f(x)=0 , 其它x顯然對於所有x, f(x)≧0又∫-∞+∞ (3/64)x(16-x² )1/2 dx = ∫04 (3/64)x(16-x² )1/2 dx= (3/64)∫04 x(16-x² )1/2 dx= (3/64)(-1/3)[(16-x² )3/2 ]04 = (-1/64)[0-(16)3/2 ]= (-1/64)(-64)= 1f為pdf(b) P(X<2)=∫-∞2 f(x) dx=∫02 (3/64)x(16-x² )1/2 dx= (-1/64)[(16-x² )3/2 ]02 = (-1/64)(123/2 - 163/2 )= -(3√3)/8 + 1= 1 - (3√3)/8≒ 0.350484.f(x)=5tanx f'(x)=(5tanx)(ln5)(sec²x) f'(π/4)=(5tan(π/4))(ln5)(sec²(π/4))=(5)(ln5)(2)=10ln5 5.令u=x1/2 u²=x, 2udu=dx ∫cos(x1/2 ) dx = 2∫ucosu du = 2(usinu-∫sinudu) (分部積分法) = 2usinu + 2cosu + c =2x1/2 sin(x1/2) + 2cos(x1/2) + c

Answer 2

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