Math 1322 Spring 1999

Exam 1

Calculator Usage: You may use any approved calculator (including the TI 81, 82, 83, and 85). Your calculator may evaluate integrals numerically but you will not receive credit for any answers that are so obtained. For example, a calculator may be used to obtain the value -3.977627090 on Problem 10. No points will be awarded for numerical answers unless the appropriate work leading up to the answer is included. For each problem, the exact answer to the question is required. Exact answers need not be decimalized. Sums may be left as sums.

1.

If and are both antiderivatives of the same function, if , , and , then what is ?

Solution

> G := x -> F(x) + C;

> eqn := G(1) = 11;

> C := solve( subs( F(1) = 2, eqn) , C );

> G(3);

> subs( F(3) = 7 , % );

2.

Evaluate .

Solution

> Int(1/x,x = 1 .. exp(1)) =
subs( x = exp(1), int(1/x,x))-subs( x = 1, int(1/x,x));

> simplify( % );

3.

If is an antiderivative of , then what is the rate of change of at ?

Solution:

> F := t -> Int( (7 + t^3)/(1 + t^2) , t);

> D(F)(2);

or

> F := t -> Int( (7 + t^3)/(1 + t^2) , t);

> F := t -> Int( (7 + t^3)/(1 + t^2) , t );

> diff( F(t) , t );

> subs( t = 2 , % );

4.

Over the course of a 2 hour period a water heater leaked at the following rates

Estimate the total leakage (in liters) by using the Midpoint Rule with 4 subintervals each of length 30 minutes.

Solution:

The plotted points are

> points_on_graph := [ [0,0] , [15,1/16] , [30 , 2/16] ,
[45,3/16] , [60,3/16] , [75,4/16] ,
[90,3/16], [105,1/16] , [120,0] ]:

> plot( [ points_on_graph ] , style = POINT, symbol = CIRCLE );

When we add the Midpoint Rule approximating rectangles we obtain:

The apploximate leakage in liters is

> (1/16*liters/minute + 3/16*liters/minute
+ 4/16*liters/minute + 1/16*liters/minute)*(30*minute);

5.

Suppose that is approximated by a Riemann sum where is the width of the subintervals and where is the right endpoint of the j th subinterval. Each summand of A may be interpreted as the area of a rectangle. What is the area of the third rectangle (counting from the left endpoint) ?

Solution:

> restart:
with(student):

> f := x -> exp(8*x^2);

> xi := seq( j*((1 - 0)/4) , j = 1 .. 4);

> Area_of_third_rectangle := f(xi[3])*(xi[3]-xi[2]);

6.

If and if is the antiderivative of for which , then what is ?

Solution:

> restart:

> f := x -> sqrt(x)*(x^3-x);

> F := x ->subs( u = x , int( f(u) , u )) + C;

> C := solve( F(0) = -2 , C );

> F(1);

7.

A particle moves along a coordinate axis in such a way that its velocity at time is given by the equation . If the coordinate of the particle is 3 at time , then what is its position at time ?

Solution:

> restart:

> v := t -> 4*t^3+t^2+1;

> p := int(v(t),t)+C;

> C := solve( 3 = subs( t = 1, p ) , C );

> subs( t = 2 , p );

8.

The area under the graph of and over the interval [0,8] is approximated by the Riemann sum where is the midpoint of the j 'th subinterval. What is the approximation?

Solution:

> restart:
with(student):

> f := x -> x/(x+1);

> middlebox( f(x), x = 0..8 , 4);

> middlesum(f(x), x = 0..8 , 4);

> value( % );

9.

If , then what is (the derivative of at 2 )?

Solution:

> F := (x) -> Int(t^3/(1+t^2),t = 1 .. x);

> D(F)(x);

> D(F)(2);

10.

Evaluate .

Solution:

> int(1/x-exp(x), x );

> subs( x = 2 ,% ) - subs( x = 1 ,% );

> simplify( % );

11.

If for in the interval [2,4] and if , then what is ?

Solution:

> restart:

> int(f(x),x = 0 .. 2) = int(f(x),x = 0 .. 4) - int(f(x),x = 2 .. 4);

> subs( int(f(x),x = 0 .. 4) = 14 , % );

> lhs( % ) = subs( f(x) = 4 , rhs( % ) );

> simplify( % );

12.

There are constants and such that is an antiderivative of . Find the constants and .

Solution:

> solve( identity( diff( a*x*ln(2*x)+b*x , x ) = ln(2*x) , x) , {a,b} );

13.

Evaluate .

Solution:

> restart:
with(student):

> J := Int(x/sqrt(x^2+1),x);

> K := changevar( u = x^2 + 1 , J , u );

> answer := subs( u = x^2 + 1 , value(K) ) + C;

14.

Suppose that where is the function whose graph appears below:

Describe the interval(s) on which is increasing.

Solution:

is increasing on the interval(s) on which its derivative, namely , is positive. Therefore is increasing on the interval (1,2) .

15.

Suppose that and are as in the preceding problem. Describe the relative extrema (maxima and minima) of .

Solution:

has a relative maximum at a point at which its graph changes over from rising to falling. At such a point the derivative of changes from positive to negative. Therefore has a local maximum at .

has a relative minimum at a point at which its graph changes over from falling to rising. At such a point the derivative of changes from negative to positive. Therefore has a local maximum at .

16.

Evaluate .

Solution:

> restart:
with(student):

> J := Int(exp(x)/(exp(x)+1),x = 0 .. b);

> K := changevar( u = exp(x) + 1 , J , u );

> value( K );

> assume( b , real);

> combine( % , ln);

17.

If , then find .

Solution:

> restart:

> A := int(D(g)(x),x = a .. b);

> g := (x) -> x^3*sin(x)/(x+1);
a := 0;
b := 1;

> A;

18.

Evaluate .

Solution:

> restart:
with(student):

> J := Int(sin(x/3+2/3),x = 0 .. b);

> K := changevar( u = x/3 + 2/3 , J , u );

> value( % );

19.

If then evaluate .

Solution:

> restart:

> A := D( x -> int( f(t) , t = v(x) .. u(x) ))(x);

> u := x -> x^2;
v := x -> x;
f := t -> sqrt( 1 + t^3 );

> A;

20.

Evaluate

Solution:

> J :=
Int(theta*cos(theta^2)*sqrt(sin(theta^2)),theta = 0 .. sqrt(Pi/2));

> with(student):

> K := changevar( u = sin(theta^2) , J , u );

> value(K);