Query3

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course MTH174

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

Optional preliminary questions:

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Question: `q001. Show by differentiation that F(x) = e^(t^3) + c is an antiderivative of f(x) = 3 t^2 e^(t^3) + c.

What is the 'inner function' of the composite e^(t^3) and how is it related to the factor 3 t^2?

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Your solution:

U = t^3

Du = 3t^2dx

Integrate e^u = e^u

Replace u with t^3

E^(t^3)

confidence rating #$&*:

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Self-critique Rating:OK

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Question: `q002. To integrate the function of the preceding problem it would be helpful to have a structured way to 'reverse engineer' the chain rule.

For integrand 3 t^2 e^(t^3):

Let u be the 'inner function' t^3.

• What is du/dt?

• What therefore is the differential du? (recall that the differential df of a function f(x) is df/dx * dx)

If you write out the integral of 3 t^3 e^(t^3) with respect to t, using standard integral notation, the expression inside the integral sign is

3 t^2 e^(t^3) dt .

How can you rewrite this expression in terms your newly created variable u without reference to the variable t? (i.e., how can you write the expression in terms of u and du rather than in terms of t and dt?)

In terms of u, what therefore is the general antiderivative of your integrand?

To express your result in terms of t, simply substitute t^3 for u. What is your result?

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Your solution:

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Question: `q003. Use the method introduced in the preceding problem to integrate cos(t) * e^(sin(t)).

The steps are:

• let u = sin(t)

• find du/dx

• find du

• write the expression cos(t) * e^(sin(t)) dt in terms of u and du

• find the general antiderivative

• substitute sin(t) for u to get your antiderivative in terms of t

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Your solution:

U = sin(t)

Du = cos(t)dt

Integrate e^u = e^u

Replace u = e^cos(t)

@&

u = sin(t)

Nothing in the process will affect the exponent in e^(sin(t)). You will never get e^cos(t).

*@

confidence rating #$&*:

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Self-critique Rating:ok

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Question: `q004. As you have seen in the preceding, the trick is to find a function u that can be used to convert the given integrand to an expression involving u rather than the original variable. This can't always be done (for example if the preceding problem had been t^2 e^(sin(t)) we would not have been able to find a function u that permitted the change of variable; we needed that cos(t) out front to accommodate the derivative of the sine function). In fact, while we can find the general antiderivative of t^2 e^(sin(t)) using other techniques, this is not possible most functions we could write down.

Each of the following expressions can be expressed in terms of u and du, provided you find the right u. See how well you can do. You need only convert the expression

• 2 t sin(t^2) dt

• t sin(t^2) dt

• sin(t) e^(cos(t)) dt

• t cos(t^2) e^(cos(t^2)) dt

• x ln(x) dx

• x^3 e^(x^2) dx

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Your solution:

• u=t^2; du = 2tdt

• u=t^2; 1/2du = tdt

• u= cos(t); -du = sin(t)dt

• u= cos(t^2);

@&

Good, but you haven't specified du.

*@

• u = x; du = 1

• u = x^2; du = 2x

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Question: `q005. Complete the integration of each of the following:

• 2 t sin(t^2) with respect to t

• t sin(t^2) with respect to t

• sin(t) e^(cos(t)) with respect to t

• t cos(t^2) e^(cos(t^2)) with respect to t

• x ln(x) with respect to x

• x^3 e^(x^2) with respect to x

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Your solution:

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Question: Section 6.5 Problem 3

6.5.3 (previously 6.5 #8) Galileo: time for unif accel object to traverse dist is same as if vel was ave of init and final; put into symbols and show why true

How can you symbolically represent the give statement?

How can we show that the statement is true?

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Your solution:

Confidence Assessment:

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Given Solution:

If acceleration is uniform, then it follows by straightforward integration of the constant a with respect to t that

v(t) = v_0 + a t,

where a is the acceleration and v_0 is the velocity at clock time t = 0.

It also follows that the position function, found by integrating the velocity function, can be expressed as

x(t) = x_0 + v_0 t + 1/2 a t^2.

From the velocity function we can get expressions for the initial and final velocity over an interval. If we average these and multiply by the duration of the interval we get a prediction fo the distance traveled by the object.

From the position function we can calculate the two positions and subtract to determine the actual change in position.

It will turn out that the two results are identical, confirming the statement:

Between clock time t = t_1 and t = t_2, the velocity will change from

v(t_1) = v_0 + a t_1

to

v(t_2) = v_0 + a t_2.

The average of these velocities is

ave of init and final vel = (v(t_1) + v(t_2) ) / 2

= ( v_0 + a t_1 + v_0 + a t_2) / 2

= v_0 + a ( t_1 + t_2 ) / 2.

The time interval is `dt = t_2 - t_1. Moving a velocity equal to the average of the two velocities the object would travel through a displacement of

`dx = (ave of velocities) * `dt

= ( v_0 + a ( t_1 + t_2 ) / 2 ) * ( t_2 - t_1)

= v_0 * (t_2 - t_1) + 1/2 a ( t_2^2 - t_1^2).

Now using the position function x(t) = x_0 + v_0 t + 1/2 a t^2:

The change in position between t_1 and t_2 is

x(t_2) - x(t_1) = (x_0 + v_0 t_2 + 1/2 a t_2^2) - (x_0 + v_0 t_1 + 1/2 a t_1^2)

= v_0 ( t_2 - t_1) + 1/2 a ( t_2^2 - t_1^2).

Our two expressions agree, proving the original statement true.

Alternative solution:

For uniform acceleration the velocity function can be expressed as

• v0 + a t.

Integrating this function between clock times t1 and t2 we obtain the displacement s:

• s = 1/2 a (t2^2 - t1^2) + v0 ( t2 - t1).

Dividing the integral (i.e., the displacement) by the length of the interval we get the average value of the function—i.e., the average velocity. The length of the interval is t2 - t1 so

• displacement / interval =

Integral / interval =

(1/2 a (t2^2 - t1^2) + v0 ( t2 - t1)) / (t2 - t1) =

1/2 a ( t2 + t1) + v0.

The initial value of the velocity on the interval is v0 + a t1, the final velocity v0 + a t2, so the average of initial and final velocities is

• Ave of init and final velocity =

(v0 + a t1 + v0 + a t2) / 2 =

v0 + 1/2 a ( t1 + t2).

This is identical to the expression obtained previously.

Thus the average of the initial and final velocities is equal to the average velocity on the interval, and the result is proven.

Still another solution:

If an object is dropped from rest and falls for time t it will reach velocity vf = a t. So the average of its initial and final velocities will be

• Ave of init and final vel =(vf + v0) / 2 = (32 t + 0) / 2 = 16 t.

The distance fallen is known to be

• s = 16 t^2.

But this is exactly the distance the object would travel in time t if its average velocity was 16 t. Thus the two quantities are identical.

A numerical example for a given s:

When s = 100, then, t is the positive solution to 100 = 16 t^2. This solution is t = 2.5.

The velocity function is v = a t. We have to find v when s = 100. When s = 100, we have t = 2.5, as just seen. So v = a t = 32 * 2.5 = 80, representing 80 ft/sec.

Now we know that s = 100, vf = 80 and v0=0. Substituting into t = s / [ (vf + v0)/2 ] we get t = 100 / [ (80 + 0) / 2 ] = 100 / 40 = 2.5.

This agrees with the t we got using s = .5 a t^2. **

Another argument:

The average value of any linear function over an interval is equal to the average of its initial and final values over the interval.

In a nutshell, because the v vs. t graph is linear, the average velocity is equal to the average of the initial and final velocities.

Since

time of fall = displacement / average velocity,

it follows that

time of fall = displacement / (ave of initial and final vel).

This latter expression is just the time that would be required to fall at a constant velocity which is equal to the average of initial and final velocities.

More rigorously:

The graph is linear, so the area beneath the graph is the area of a triangle.

The base of the triangle is the time of fall, and its altitude is the final velocity.

By a simple construction we know that the area of the triangle is equal to the area of a rectangle whose length is equal to the base of the triangle, and whose width is equal to half the altitude of the triangle.

It follows that the area of the triangle is equal to half the final velocity multiplied by the time interval.

Since the initial velocity is 0, the average of initial and final velocities is half the final velocity.

So the area of the triangle is the product of the time of fall and the average of initial and final velocities

area beneath graph = time of fall * ave of init and final vel

The area beneath the graph is equal to the integral of the velocity function over the time interval, which we know is equal to the displacement. So we have

displacement = time of fall * ave of init and final vel, so that

time of fall = displacement / ave of init and final vel.

This leads to the same conclusion as above.

The same argument, expressed more symbolically:

A graph of v vs. t graph is linear. Over any time interval t = t1 to t = t1 + `dt, displacement is represented by the area of the corresponding trapezoid, which is d = (v1 + v2) / 2 * `dt.

Solving for `dt we again obtain `dt = d / [ (v1+v2)/2 ].

Still another way of structuring the argument:

** Using s for the distance fallen we can translate Galileo's statement as follows:

• t = s / [ (vf + v0)/2 ].

An object accelerating for time t, starting with initial velocity v0 at t = 0, has velocity function

• v = v0 + a * t

and position function

• s = .5 a t^2 + v0 t,

assuming that s = 0 at t = 0.

Both of these results are easily obtained, for the given initial condition that initial velocity is zero, by a straightforward integration of the acceleration function (to get the velocity function) then the velocity function (to get the position function).

For given displacement s we can solve the position equation for t. The equation is rearranged to the form of a standard quadratic:

.5 a t^2 + v0 t - s = 0, with solutions

• t = (-v0 +- sqrt(v0^2 + 2 a s) ) / a ).

Substituting this into the velocity function we obtain the final velocity:

Final velocity

= v0 + a t = v0 + a * (-v0 +- sqrt(v0^2 + 2 a s) / a)

= +- sqrt(v0^2 + 2 a s) .

The average of the initial and final velocities is therefore

• ave. of init and final vel = (initial vel + final vel) / 2 = (v0 +- sqrt(v0^2 + 2 a s)) / 2.

Traveling at this velocity for time t = (-v0 +- sqrt(v0^2 + 2 a s) ) / a ) the displacement will be

• Displacement = average velocity * time of travel

= ( v0 +- sqrt(v0^2 + 2 a s)) (-v0 +- sqrt(v0^2 + 2 a s) ) / a ) / 2

The numerator is the product of the sum and the difference of two quantities and simplifies to 2 a s. The denominator is 2 a so the expression simplifies to just s.

This confirms that the distance traveled is the same as the distance that would be traveled at the average of the initial and final velocities.

Self-critique

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Question: **** query problem 7.1.18 integral of ` sqrt(cos(3t) ) * sin(3t)

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Your solution:

U = cos3t

-du = sin3t dt

(u^(1/2+2/2))/(3/2)

(2/3)(cos3t)^(3/2)

confidence rating #$&*: 3

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Given Solution:

** TYPICAL INCORRECT SOLUTION: (-2/3) (cos3t)^(3/2)

INSTRUCTOR COMMENT: The derivative of cos(3t) is -3 sin(3t), wo the derivative of -2/3 (cos(3t))^(-3/2) is 3 sqrt(cos(3t) sin(3t). This solution is not consistent with the Chain Rule.

To perform the integral use substitution. Very often the first substitution you want to try involves the inner function of a composite, and that is the case here. Cos(3t) is the inner function of the composite sqrt(cos(3t)), so we try using this as our substitution:

w = cos (3t) dw = -3 sin (3t)

so that sin(3t) = -dw / 3.

Thus our expression becomes

w^(1/2) * (-dw / 3).

The integral of -1/3 w^(3/2) with respect to w is -1/3 * (2/3) w^(3/2), treating w^(1/2) as a power function.

This simplifies to

-2/9 w^(3/2) or

-2/9 * (cos(3t))^(3/2).

The general antiderivative is

-2/9 * (cos(3t))^(3/2) + c,

where c is an arbitrary constant.**

DER

22:16:28

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Self-critique (if necessary): I didn’t use the chain rule

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Self-critique Rating: 3

@&

Note that the derivative of your result is not the original function.

You always need to check your integration by differenting your result and comparing with the original integrand.

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Question: Section 7.1 problem 6

7.1.6 (previously 7.1.21) antiderivative of x^2 e^(x^3+1)

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Your solution:

U = (x^3) + 1

Du = 3x^2

Du/3 = (x^2)dx

(E^(x^3)+1)/3 + C

confidence rating #$&*: 3

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Given Solution:

Substituting for the inner function of the composite:

• u = x^3 + 1 yields

• du = 3 x^2 dx, so

• x^2 dx = 1/3 du.

This makes the integrand 1/3 e^u du.

An antiderivative is then

• antiderivative = 1/3 e^u, or 1/3 e^(x^3+1)

leading to the general indefinite integral

• indefinite integral = 1/3 e^(x^3+1) + c

If we take the derivative of this function we do indeed obtain our original integrand.

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Self-critique (if necessary):OK

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Self-critique Rating:OK

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Question: Section 7.1 Problem 9

7.1.9 (previously 7.1.35) Find an antiderivative of (t+1)^2 / t^2

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Your solution:

(t + 1)^2 = t^2 + 2t + 1

(t^2 + 2t + 1)/ (t^2)

1 + 2t^(-1) + t^(-2)

[(1t^(0+1))/1] + 2ln(t) + (t^(-2 + 1))/-1

t^1 + 2ln(t) - (1/t)

confidence rating #$&*: 3

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Given Solution:

expand (t+1)^2 to get t^2+2t+1.

divide by t^2 to get 1 + 2t^-1 + t^-2.

Each term of this function is a power function. Integrate term-by-term and add the integration constant to get the general antiderivative

• general antiderivative = t + 2ln(t) - t^-1 +C

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Self-critique (if necessary): OK

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Self-critique Rating:

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Question: Section 7.1 Problem 13

7.1.13 (previously 7.1.60). int(1/(t+7)^2, t, 1, 3) **** What did you get for the definite integral?

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Your solution:

U = t + 7

Du = 1dt

1/u^2

(U^-2+1)/-1

(U^-1)/-1

-1/u

-1/(t + 7)

-1/(3 + 7) - -1/(1+7)

1/40

confidence rating #$&*: 3

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Given Solution:

This situation involves a composite of the power function 1 / z^2 and the linear function t + 7. The latter is the ‘inner’ function of the composite.

We therefore substitute u = t+7 to get du = dt; limits t = 1 and t = 3 become u = 8 and u = 10. So the integral becomes

integral( u^-2, u from 8 to 10).

The integrand is the power function u^-2, or 1 / u^2. Its simplest antiderivative can be expressed as -u^-1 or -1/u.

So the definite integral is

• change in antiderivative =

-1/10 - (-1/8) =

1/8 - 1/10 =

1/40.

Amplified explanation

The given definite integral becomes the integral of 1 / u^2, integrated from u = 8 to u = 10. We can find our integral just in terms of the function u:

The definite integral is the change in the antiderivative function. The antiderivative function is - 1 / u, which between u = 8 and u = 10 changes from -1/8 to -1/10.

Subtracting -1/8 from -1/10 we get 1/40.

Alternatively, having found our antiderivative in terms of u, we could have substituted t + 7 for u and done the integral in terms of t:

Our antiderivative -1/u is equal to -1 / (t + 7).

Our original integral was form t = 1 to t = 3.

At t = 1 our antiderivative is -1/(1 + 7) = -1/8. At t = 3 our antiderivative is -1 / (3 + 7) = -1/10. Once more the change in the antiderivative is found to be 1/40.

Still another way of putting it:

If f(t) = 1 / (t + 7)^2, our antiderivative is F(t) = - 1 / (t + 7). Thus

integral ( f(t), t from 1 to 3) = F(3) - F(1) = -1/10 - (-1/8) = 1/40.

F(3) - F(1) is, of course, the change in the value of F(t) between t = 1 and t = 3; i.e., this is the change in the value of the antiderivative function.

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Self-critique (if necessary):OK

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Question: Section 7.1 Problem 16

7.1.16 (previously 7.1.86). World population P(t) = 5.3 e^(0.014 t).

What were the populations in 1990 and 2000?

What is the average population between during the 1990's and how did you find it?

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Your solution:

1990 = 5.3

2000 = 6.09

(5.3 + 6.09)/2 = average population during 1990s

5.70

confidence rating #$&*:2

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Given Solution:

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In 1990, t = 0 so the population is P(0) = 5.3 billion

In 2000, t = 0 so the population is P(10) = 6.1 billion

To find the average population we integrate the population function over the 10-year interval, then divide by the 10 years:

An antiderivative of 5.3 e^(0.014 t) is found by letting u equal the ‘inner function’ of the composite:

u = .014 t so

du = .014 dt and

dt = du / .014.

Thus the expression 5.3 e^(0.014 t) dt becomes 5.3 e^u * du / .014.

An antiderivative of e^u with respect to u is just e^u, so the antiderivative of the function is

• 5.3 e^u / .014 = 380 e^u, approximately. In terms of t this is

• 380 e^(.014 t).

At t = 0 this antiderivative would have value about 380.

At t = 0 your antiderivative would have value about 435.

• The change in the value of the antiderivative is therefore about 435 - 380 = 55.

• The change in the antiderivative is the definite integral of the function.

The average value of the function is equal to its definite integral divided by the time interval:

• average value = 56.89 / 10 = 5.689.

This is quite close to, but a little less than the average of the initial and final populations.

It is less because the exponential function is concave upward, and the population curve therefore 'dips' below the straight line connecting the initial and final points.

However the curvature on this interval is not pronounced, and the function never 'dips' far below the straight line, so the average is close to the straight-line average you gave in your initial solution.

INCORRECT IF NOT UNREASONABLE ANSWER:

The average population seems as if it would be the average of the two, therefore

6.1 + 5.3/ 2 = 5.7 billion

INSTRUCTOR RESPONSE:

Review the definition of the average value of a function:

• The average value of a function over an interval is equal to its integral over that interval, divided by the length of the interval.

Now if the function is linear, then the average of its initial and final values is equal to its average value (see the above given solution to the Galileo problem). However:

• If the function is not linear, it is very unlikely that the average of its initial and final values is equal to its average value

• If the function has nonzero positive or negative concavity on the entire interval it will never have an average value equal to the average of its initial and final values.

The exponential function is not linear, and is in fact concave upward on this interval, so averaging initial and final values on the interval won't work here.

You have to integrate the function and divide by the 10-year interval of integration.

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COMMON INTEGRATION ERRORS: integral( x^2 * e^(x^3+1) ) = integral( x^2) * integral (e^(x^3 + 1) ) = x^3 / 3 * e^(x^3 + 1)

INSTRUCTOR RESPONSE: You have essentially assumed that the integral of a product function is equal to the product of the integrals. The integral is of the form

• integral (f * g)

with f(x) = x^2 and g(x) = e^(x^3 + 1).

However integral (f * g) is not the same as integral (f) * integral(g):

The derivative of our antiderivative function must equal the original integrand. That is, the derivative of integral(f * g) must be f * g.

However the derivative of integral (f) * integral(g) is expressed by the product rule as

(integral (f) * integral(g)) ' = ( integral (f) ) ' * integral(g) + integral (f) * (integral(g)) ' = f * integral(g) + g * integral(f).

This is much different than the original integrand f * g, so integral ( f * g) is not generally the same as integral(f) * integral(g).

Without going into the details, we'll also assert that

• integral(f / g) is not the same as integral(f) / integral(g),

which you can if you wish verify for yourself by taking the derivative of integral(f) / integral(g), using the quotient rule. What you get is nothing like the original integrand f / g.

So for example integral( (t+1)^2 / t^2 ) is not the same as integral ( t + 1) / integral (t^2).

Note also that, while the integral of e^x with respect to x is e^x + c, the integral of (e^(x^3 + 1) ) with respect to x is NOT the same as e^(x^3 + 1) + c:

• The derivative of e^(x^3 + 1) is 3 x^2 * e^(x^3 + 1), not e^(x^3 + 1).

So e^(x^3 + 1) is not an antiderivative of e^(x^3 + 1).

Self-critique

Needed to integrate to find the average

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Self-critique rating: 3

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Question: `q006. Integrate each of the following by the method of substitution:

• t sin(3 t^2 + 2) with respect to t

• sin(t) e^(2 pi cos(t)) with respect to t

• t cos(t^2 + 1) e^(cos(t^2 + 1)) with respect to t

• x ln(x+1) with respect to x

• x^3 e^(x^2 - 3) with respect to x

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Your solution:

• u=3t^2 + 2; du = 6t; du/6 = t dt; (-cos(3t^2+2))/6

• u=2pi cost; du = -2pi sinx; -du/2pi = sinxdx; -(e^(2pi cost))/2pi

• u = cos(t^2 + 1); du = -2t sin(t^2 +1); -du/2 = tsin(t^2 +1)dt;

• u = x+1; du = 1; int: x ln(u); 1 ln(u) + 1/u * x; ln(x+1) + x/(x+1)

• u = x^2 -3; du = 2x; du/2x = dx; int: x^3 e^u; 3x^2 (e^u) (e^u); 3x^2 e^2(x^2 - 3)

@&

x^2 = u + 3, so the 3 x^2 will be expressed as 3 (u +3).

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&#Your work looks good. See my notes. Let me know if you have any questions. &#