How to calculate displacement when acceleration and time are known

How to calculate displacement when acceleration and time are known

Known: acceleration a, time t
Solution: displacement s
If the object is moving in a straight line with constant speed, it can be calculated according to the following formula
S = v0t + at ^ 2 / 2. When the original speed V0 = 0, s = at ^ 2 / 2

When an object is moving in a straight line with uniform acceleration, the initial velocity is 0.5m/s, and the displacement in the 7th s is 4m more than that in the 5th s, the following results can be obtained: (1) the acceleration of the object; (2) the displacement of the object in the 5th s

(1) Using the formula of displacement difference in adjacent equal time: △ x = at2, it is obtained that: a = X7 − x52t2 = 42 = 2m / S2 (2) the displacement of the object in 5S is: x = v0t + 12at2 = 0.5 × 5 + 12 × 2 × 25 = 27.5m A: (1) the acceleration of the object is 2m / S2; (2) the displacement of the object in 5S is 27.5m

When an object moves in a straight line with uniform speed change, the scale size is 4 m / s at a certain time, and the speed becomes 10 m / s after 1 s
A. The displacement may be less than 4MB. The displacement may be greater than 10mc. The acceleration may be less than 4m / S2D. The acceleration cannot be greater than 4m / S2

If the initial velocity direction of the object is positive, then V0 = 4m / s, and the final velocity v = ± 10m / s. according to the definition of acceleration, a = V − v0t has: when the final velocity is positive, the acceleration A1 = 6m / S2; when the final velocity is negative, the acceleration A2 = − 14m / S2, so CD is wrong

Finding the indefinite integral of X (arctanx)

∫x(arctanx)dx
=(1/2)∫ (arctanx)d(x^2)
= (1/2)x^2(arctanx) -(1/2)∫ x^2 (1/(1+x^2) dx
= (1/2)x^2(arctanx) - (1/2)∫ dx+ (1/2)∫ 1/(1+x^2) dx
= (1/2)x^2(arctanx) - (1/2)x + (1/2) arctanx + C

Finding indefinite integral ∫ x ^ 2 * arctanx

Indefinite integral of (x ^ 2 + 3) arctanx

The original formula = ∫ x ^ 2arctanxdx + 3 ∫ arctanxdx integrates the two parts separately, ∫ x ^ 2arctanxdx = ∫ arctanxd (x ^ 3 / 3) = (x ^ 3 / 3) arctanx - (1 / 3) ∫ x ^ 3D (arctanx) = (x ^ 3 / 3) arctanx - (1 / 3) ∫ x ^ 3DX / (1 + x ^ 2) = (x ^ 3 / 3) arctanx - (1 / 3) ∫ (x ^ 3 + x-x) DX / (1 +

Urgently seeking indefinite integral of 3 ^ Xe ^ xdx

Dizzy, the landlord really willing to give points
According to the formula, [(3e) ^ x] '= (3e) ^ x * ln (3e) = (3e) ^ x * (1 + Ln3) DX
So the integral of (3e) ^ x * DX = (3e) ^ X / (1 + Ln3)

Indefinite integral ∫ (Xe ^ x) / (1 + x) ^ 2DX

∫ Xe ^ X / (1 + x) ^ 2 DX = ∫ [e ^ x (1 + x) - e ^ x] / (1 + x) ^ 2 DX = ∫ e ^ X / (1 + x) DX - ∫ e ^ X / (1 + x) ^ 2 DX = ∫ e ^ X / (1 + x) DX - ∫ e ^ x D [- 1 / (1 + x)] = ∫ e ^ X / (1 + x) DX + e ^ X / (1 + x) -∫ 1 / (1 + x) d (e ^ x), points

Indefinite integral of∫ xe∧ - x∧2dx
Please write down the process of solving the problem

∫xe∧-x∧2dx
=（-1/2）∫e∧-x∧2d(-x²)
=(-1/2)e^(-x²)+C

Indefinite integral of Xe ^ - x ^ 2cote ^ - x ^ 2DX

∫xe^(-x^2)cote^(-x^2)dx
= -(1/2)∫e^(-x^2)cote^(-x^2)d(-x^2)
= -(1/2)∫cose^(-x^2)/sin[e^(-x^2)]d[e^(-x^2)]
= -(1/2)∫1/sin[e^(-x^2)]d[sine^(-x^2)]
= -(1/2)ln|sine^(-x^2)|+C