Elementary Calculus Questions, Introductory Calculus Questions

Let the real-time velocity be v(t).

Then v'(t)=kv(t)

It's a differential equation, it's easy to solve.

By the initial value v(0)=vo

v（t）=vo-1+e^（kt）

Let v(t)=0 to get the required time.

The lower limit for definite integration of v(t) is vo and the upper limit is 0

What is obtained is the elapsed displacement.

By: xom723 - 4-4 15:58v=v.

e to the power -kt, when t tends to infinity, v is equal to 0, so theoretically the velocity can never be 0; The total distance traveled is v. /k.Solution:

dv dt = divide by the past, multiply dt, and then from v to v. To the v integral, t from 0 to t integral, v=v. *e to the power of -kt, and then integrate both sides of t from 0 to infinity to obtain s=v.

t.Author: xxh40089310 - Probationary period Level 1 4-4 16:12kIf the positive number is taken a=-kv=dv dt 1 v dv=-k dt The two sides are integrated to obtain lnv-lnv0=-kt v=v0*e (-kt)When v approaches 0, t is infinite, so time is infinite.

and v=ds dt, so ds=v0*e (-kt) dt is then integrated to get s=-v0 k*[e (-kt)-1], and time t is close to infinity, so the displacement s=v0 k

Two important limits and an equivalent infinitesimal substitution are sufficient.

If k is taken as a positive number, a=-kv=dv dt 1 v dv=-k dt The integration of both sides gives lnv-lnv0=-kt v=v0*e (-kt) When v approaches 0, t is infinity, so time is infinity. and v=ds dt, so ds=v0*e (-kt) dt is then integrated to get s=-v0 k*[e (-kt)-1], and time t is close to infinity, so the displacement s=v0 k

Most of the problems can be found by using Lopida's rule, that is, the numerator and denominator are derived at the same time, and there are unknown numbers on the index, use ln to divide the exponent, and if there is something at night, the specific ...... will not be solved

=limn*[√n^2+1) -n]

lim n*[ n 2+1)-n][ n 2+1)+n] [ (n 2+1)+n] Note: The numerator and denominator are multiplied by (n 2+1)+n

lim n*[n^2+1-n^2]/[√(n^2+1)+n]=lim 2n/[√(n^2+1)+n]

lim 2 [ (1+1 n 2) +1] Note: The numerator and denominator are divided by n

lim 2/[√(1+0) +1]=1

Split term: (1 2+x ) (1+x )=[(x +1)-1+1 2] (x +1)=1-1 [2·( x +1)]

Original integral = [1-1 2(x +1)]dx=x-1 2·arctanx+c, c is a constant (where, dx(x +1)=arctanx+c)ps: I hope my answer is helpful to you.

Don't ask for an additional 50, as long as you adopt it in time!

This is the definition of a definite integral.

The area of the graph formed by f(x) on [a,b] and the x-axis is approximately [f(x)δx].

When δx->0, i.e., the definite integral = limδx->0 [f(x)δx] is now the interval [1,5].

Then the upper and lower limits of the integral b=5 and a=1

The integral function is f(x)e x x

Solution: Profit Prof(Q) = Total Revenue - Total Cost = R(Q)-C(Q), R(Q) = (MR)DQ= (100-5Q)DQ=, obviously, when Q=0, R(0)=0, C(0)=100, C1=0, C2=100. ∴prof(q)=-12q+。

Again, the profit function prof(q) is the derivative of q and its value is 0, and there is [prof(q)].'=mr-mc=-q²+13q-12=0。The extreme points of prof(q) are q1=1 and q2=12.

However, prof(q1)<0, prof(q2)=116, when q=12, the profit is the largest, and its value is 116 (10,000 yuan).

FYI.

Defined by definite integrals, there are a=1, b=5, f(x)=(e x) (1+x).

FYI.

5. Select A

Option A cosx-1 is equivalent to , is the higher order infinitesimal of x, b option x+x = x (x+1) is equivalent to x, is the same order infinitesimal of x, c option sinx is equivalent to x, and d option x is the lower order infinitesimal of x.

6. Select b as 0 as the first derivative and 0 as the maximum value < the second derivative.

The second derivative >0 is the minima.

Here f'(a) = 0, while f''(a) = -4<0, so x=a is the maximum point of f(x).

So choose B7 and choose B

The left limit of f(x) = 3, while the right limit = 1, and the left and right limits of f(1) = 3 are not equal, then it must be discontinuous.

The function is discontinuous at this point, and it is impossible for the left derivative and the right derivative to exist at the same time, the left derivative = 6, and the right derivative = lim(dx->0+) f(1+dx) -f(1)] dx

lim(dx->0+) 1+dx) 3 -3] dx=lim(dx->0+) dx) 2 +3dx +3 -2 dx tends to negative infinity, i.e. the right derivative does not exist, so choose b

x=y=15+

Isn't it enough to get this list of areas and lengths to find the differential?

First of all, f(x)=x 5-3x-1 is a continuous function, and f(1)=-3<0, f(5)=5 5-3*5-1>0;Thus, according to the existence theorem of roots, the equation x 5-3x-1 = 0 has at least one real root between (1,5).

Answer: Let f(t)=t(1-2t)(1-3t) t [0,1].

It is advisable to let f(t)=t(1-2t)(1-3t) a(3t-1) be established in [0,1] constantly, and determine a first

Because the inequality sought is equal at x=y=z=1 3.

Therefore, f(t)=t(1-2t)(1-3t)-a(3t-1) takes the minimum value when t=1 3, and the derivative is 0

Therefore 18t 2-10t+1-3a=0 has a root of x1=1 3, so x2=2 9, a=25 81

So g(t) increases monotonically at [0,2 9], [1 3,1] and decreases monotonically at [2 9,1 3].

So the minimum value of g(t) on [0,1] is min=0

So x(1-2x)(1-3x)+y(1-2y)(1-3y)+z(1-2z)(1-3z).

f(x)+f(y)+f(z)

25(3x-1)/81+25(3y-1)/81+25(3z-1)/81=0

Take etc if and only if x=y=z=1 3.

f(x)=2x^3-9ax^2+12a^2x

a=1, then there is f(x)=2x 3-9x 2+12x, f'(x)=6x^2-18x+12=6(x^2-3x+2)=6(x-1)*(x-2)

The tangent equation for the origin is y=kxIf the tangent coordinates are (xo,yo), then there is k=yo xo=6(xo 2-3xo+2).

yo=6(xo^3-3xo^2+2xo)=f(xo)=2xo^3-9xo^2+12xo

The solution yields 4xo 3-9xo 2=0

xo^2(4xo-9)=0

xo=0(rounded), xo=9 4

yo=2*9^3/64-9*9^2/16+12*9/4=27-729/32=135/32

Therefore, the tangent coordinates are (9 4,135 32).

So the tangent equation is y=135 72 x

x)=6x^2-18ax+12a^2=6(x-a)(x-2a)=0

Get x1=a, x2=2a

a>0, then there is f'(x) >0, function increases, in a< x<2a, f'(x) <0, function minus.