Find ∫ lower limit-3 upper limit 3 (2x + 3 + 13-2x) DX. Content of mathematical definite integral and calculus in higher 2

Find ∫ lower limit-3 upper limit 3 (2x + 3 + 13-2x) DX. Content of mathematical definite integral and calculus in higher 2

Better song

Y = 2x + 3x when x equals 2 to 2.01 calculus

Are you wrong?
y=5x?
The solution is: dy = 5DX, so dy / DX = 5, that is, y '= 5
Integral: take integral on both sides, left sydy = 2.5 s x ^ 2 DX
=(5/2)*((2.01)^2-2^2)
That's it. S is the integral sign

Y = (1 + x ^ 2) arctanx calculus

Differential and integral are two different operations. There are calculus courses, but none of them is called calculus. I estimate how many points you need ∫ (1 + x ^ 2) arctanx DX = ∫ arctanxd (x + x ^ 3 / 3 + C) = (x + x ^ 3 / 3 + C) arctanx - ∫ (x + x ^ 3 / 3) darctanx = (x + x ^ 3 / 3 + C) arctanx - ∫ (x + x ^ 3 /

For the problem of calculus, the tangent on the curve y = x ^ 2-4x + 1 is parallel to the straight line, and the coordinates of the point y = 2x + 1 are () Such as the title One word is missing: The tangent on the curve y = x ^ 2-4x + 1 is parallel to the straight line, and the X coordinate of the point y = 2x + 1 is (), Is the coordinate of x 1 / 2? Why doesn't it feel right? I have four alternative answers: (a) 3 (b) - 3 (c) 1 (d) - 1

y'=2x-4=2,x=3
This doesn't need calculus

Find the graphic area enclosed by y = x ^ 3, y = - x ^ 2 + X + 1, senior two mathematics, calculus, process and speed

Simultaneous y = x ^ 3, y = - x ^ 2 + X + 1, the solution is: x = - 1, x = 1,
Therefore, the integral interval is: [- 1,1],
In the [- 1,1] curve y = - x ^ 2 + X + 1, which is higher than the curve y = x ^ 3,
So the enclosed graphic area
=∫[-1,1]（-x^2+x+1-x^3）dx=(-1/3*x^3+2x^2+x-1/4*x^4) |[-1,1]
=29/12-13/12=4/3.

Calculus ∫ DX / ((1-x ^ 4) ^ (- 1 / 2)) =?

∫ DX / ((1-x ^ 4) ^ (- 1 / 2)) = ∫ (1-x ^ 4) ^ (1 / 2) DX let x = Sint original formula = fcostdsint = f (cost) ^ 2dcost = (cost) ^ 3 / 3 + C = (cosarcsint) ^ 3 + C