Test 2
2008, Fall

Investigation of functions. 15 points.

Do the total examination of the function (domain, intercepts, increasing/decreasing parts, minima/maxima, convex/concave parts, inflexions, graph of the function) : (x^2) (e^(-x))

Elementary antiderivatives. 15+5 points.

Find the indefinite integral of the given function using fundamental integration. When necessary, first modify / simplify the function using identities to transform it into a linear combination of fundamental integrals!

∫((((x^2)^(1/5))-(x^3)(2^x))/(x^3))dx ,                          ∫ ((x+1)/(x (1-x)))dx.

Integration by parts and substitution. 15 points.

Integrate the function using the method of substitution or by parts.

∫(e^x) (x- (1/(1+(e^x))) )dx

Graphical antidifferentiation. 5+10 points.

The figure below shows the derivative of e function F(x).

a) Where does F(x) have maxima, minima and inflection points?

b) Plot the elements of the tangent field at the points that you think necessary. Draw the graphs of the possible functions F(x).

Plot[Evaluate[(((x-1))^2)(x)  (x+1)],{x,-1.2,1.5},AspectRatio->Automatic,AxesLabel->{x,y}]

Definite integral. 5 points.

Calculate the integral mean of the following funcion on the interval [-4,5].

f[x_]:=Which[x≤-3,3,x≤-1,-x,x≤1,0.5x+1.5,x≤2,2,x≤3,6-2x,True,0]-1
Plot[-f[x],{x,-4,4},PlotStyle->{Thickness[0.01]},Axes->True,AspectRatio->Automatic,GridLines->{{-4,-3,-2,-1,0,1,2,3,4,5},{-4,-3,-2,-1,0,1,2,3,4,5}},Ticks->Automatic,PlotRange->All]

Definite integral. 5 points.

Calculate the following integral.

(∫_(-1)^2)(1/(x+2))dx

Area. 10 points.

Calculate the absolute area enclosed by the graphs of the following functions:

f(x)=2x+1 és g(x)=1-2x+(x^2)

Variation of a function. 20 points.

The rate of change of the concentration c(t) of a material over time t  is given by the function c'(t)=v(t)=-(t/(4(((1+(t^2)))^2)))(g/s).

a) Find the antiderivative of  v(t). (10 pts)

b) Calculate the change of the concentration over the interval. (5 pts)

c) Determine the formula for c(t) under the initial condition c(0)=1. (5 pts)

Investigation of functions. 15 points.

Do the total examination of the function (domain, intercepts, increasing/decreasing parts, minima/maxima, convex/concave parts, inflexions, graph of the function) : f(x)=x (e^(-(x^2)))

Elementary antiderivatives. 15+5 points.

Find the indefinite integral of the given function using fundamental integration. When necessary, first modify / simplify the function using identities to transform it into a linear combination of fundamental integrals!

∫((2x(x^(1/3))-x)/(x^2))dx ,                          ∫ ((1-x)/(x (3-x)))dx.

Integration by parts and substitution. 15 points.

Integrate the function using the method of substitution or by parts.

∫ln(x) ((1/x)- (x^2) )dx

Graphical antidifferentiation. 5+10 points.

The figure below shows the derivative of e function F(x).

a) Where does F(x) have maxima, minima and inflection points?

b) Plot the elements of the tangent field at the points that you think necessary. Draw the graphs of the possible functions F(x).

Plot[Evaluate[-(((x+1))^2)(x)  (x-1)],{x,-1.5,1.2},AspectRatio->Automatic,AxesLabel->{x,y}]

Definite integral. 5 points.

Calculate the integral mean of the following funcion on the interval [-4,5].

f[x_]:=Which[x≤-3,3,x≤-1,-x,x≤1,0.5x+1.5,x≤2,2,x≤3,6-2x,True,0]-1
Plot[f[x],{x,-4,4},PlotStyle->{Thickness[0.01]},Axes->True,AspectRatio->Automatic,GridLines->{{-4,-3,-2,-1,0,1,2,3,4,5},{-4,-3,-2,-1,0,1,2,3,4,5}},Ticks->Automatic,PlotRange->All]

Definite integral. 5 points.

Calculate the following integral.

(∫_(-1)^2)(((x-2))^2)dx

Area. 10 points.

Calculate the absolute area enclosed by the graphs of the following functions:

f(x)=4x+1 és g(x)=1+2x+(x^2)

Variation of a function. 20 points.

The rate of change of the concentration c(t) of a material over time t  is given by the function c'(t)=v(t)= (t^2) (e^(-t))(g/s)

a) Find the antiderivative of  v(t). (10 pts)

b) Calculate the change of the concentration over the interval [0, 2]. (5 pts)

c) Determine the formula for c(t) under the initial condition c(0)=1. (5 pts)