course Mth 174
This is the practice test for the first test. Sorry it took so long to get it to you but i just started all over with a new one. Just give me as much information on each question as you can cause it would help me out along.Thanks
You didn't get most of these. See my notes.
In order to be prepared for this test, as I've been suggesting in your work, you need to have provided more detail in your solutions and self-critiques. I can't see clearly what you've been thinking. I have cautioned you about checking antiderivatives by taking the derivatives, a practice that applied to the assigned problems would have great benefit and would guide you in your choice of integration techniques.
I don't see any use of substitution or integration by parts on any of these problems, and one or another of these techniques is needed on almost every problem.
See also my notes on the principles and the details of constructing antiderivative graphs.
I'll be glad to answer subsequent questions on these notes.
Problem Number 1
Sketch a graph of a continuous function f(x) which is linear from (0, 2) to ( 2, -3.001), then linear to ( 10, 7) and then again linear to ( 14, 2). Sketch a graph of its antiderivative F(x) for which F(10)= 18 and label the known points on this graph.
I guess I will just have to describe these graphs. For the first continuous graph of function f(x) it is linear from (0,2) to (2,-3.001) and then to (10,7) and then to (14,2). It passes through the x-axis at (5.5,0).
The second graph is going to be hard because it starts off at (0,0) and then it goes up to (1,2) which is a critical point, then comes back down and crosses the x-axis at (2,0). After that it does down to another critical point (5.5,-3) and turns back up. It crosses the x-axis again at (10,0) and it ends up going up to (14,2) which is another critical point. There are inflection points at (.5,1.5), (3,-2), and (10,0).
F(x) is an antiderivative of f(x), so that f(x) is the derivative of F(x). Thus the slope of F(x) at any point is equal to the value of f(x) at that point. On any interval the area beneath the f(x) curve is equal to the change in F(x), with the proviso that areas beneath the x axis are regarded as negative. When f(x) = 0 the function F(x) has 0 slope and therefore a critical point.
f(x) is 0 at x = .8 (2/5 of the way between x = 0 and x = 2). The region between x = 0 and x = .8 is a triangle with altitude 2 and width .8, so its area is .8. Thus between x = 0 and x = .8 the antiderivative function F(x) changes by +.8. The slope of the antiderivative function F(x) is equal to the value of the function f(x), so on this interval the slope starts out at 2 (when x = 0) and decreases to 0 when x = .8. So x = .8 is a critical point of F(x).
The slopes of the F(x) graph are equal to the values of the f(x) graph, so every point where f(x) = 0 the F(x) function has a critical point. Where f(x) is positive F(x) is increasing, where f(x) is negative F(x) is decreasing.
Between x = .8 and x = 2 the graph of f(x) forms a triangle below the x axis, having altitude 3 and width 1.2 so its area is 1.8. This region is below the x axis, so the antiderivative changes by -1.8. At this point the antiderivative has decreased by -1 relative to its x = 0 value.
Similar analysis yields the following:
The linear function from (2, -3) to (10, 7) reaches 0 at x = 4.4; between x = 2 and x = 4.4 the graph forms a triangle below the x axis with area 3.6, so the antiderivative changes by another -3.6. x = 4.4 is a critical point; F(x) is decreasing to the left and increasing to the right of this critical point.
Between x = 2 and x = 4.4 the function f(x) is negative and increasing toward zero so so F(x) is decreasing at a decreasing rate; the rate of change of F(x) reaches 0 at x = 4.4; between x = 4.4 and x = 10 the function f(x) is positive and increasing so F(x) increases at an increasing rate.
Between x = 4.4 and x = 10 the graph area is 1/2 * 5.6 * 3.5 = 9.8 so the antiderivative changes by +9.8 on this interval.
Between (10, 7) and (14, 2) the graph forms a trapezoid with width 4 and average 'altitidue' 4.5, so its area is 18 and the antiderivative changes on this interval by another +18 units. On this interval f(x) is positive so that F(x) is increasing, with f(x) decreasing so that F(x) increases at a decreasing rate.
The net change of F(x) from x = 0 to x = 10 is .8 - 1.8 - 3.6 + 9.8 = +5.2. Since F(10) = 18, this means that F(0) must have been 18 - 5.2 = 12.8.
We conclude that F(x) begins at x = 0 with value 12.8, increases by .8, increasing at a decreasing rate to the x = .8 value of 13.6, then decreases at an increasing rate to the x = 2 value 11.8, then decreases at a decreasing rate to value 8.2 at x = 4.4, then increases at an increasing rate to value 18 at x = 10 and finally increases at a decreasing rate to value 36 at x = 14.
.Problem Number 2
Sketch a graph of a smooth curve through points (0, 4) to ( 4, -5.001), ( 9, 8) and ( 14, 2). Assuming that this graph represents the function f '', sketch graphs of f ' and f, assuming that both graphs pass through the origin.
These graphs are to hard to explain but I do know how to graph them and take the derivatives of them and the antiderivatives so if there are any other pointers here you could give me just let me know.
f ' is an antiderivative of f '', so to get a graph of f ' you would apply the principles of the preceding problem to the graph of f ''. It isn't expected that you calculate all the exact values of the areas, etc., but you should start your graph of f ' at x = 0 with a slope of 4, your slope decrease to 0 then to -5 by the time x = 4, then should increase first to 0 and to 8 by the time x = 8, and should then decrease to 2 by the time x = 14.
Then you proceed similarly to sketch a reasonable graph of f. The slopes of your f graph should equal the slopes of your f ' graph.
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.Problem Number 3
Find the general antiderivative of x^ 8 cos(x^ 9) / sin(x^ 9).
x^8 cosx^9 / sinx^9 = (1/9)x^9 * -sinx^9 * (1/10)x^10 / cosx^9 * (1/10)x^10 = (1/9)x^9 * -sinx^9 / cosx^9 = (1/9)x^9 – tanx^9
The derivaive of 1/9 x^9 - tan (x^9) is not x^8 cos(x^9) / sin(x^9).
Make the substitution u = sin(x^9) so that du = 9 x^8 cos(x^9) dx. Your integrand x^ 8 cos(x^ 9) / sin(x^ 9) becomes 1/9 * 1/u du with antiderivative 1/9 ln | u | + c = 1/9 ln (sin(x^9) ) + c.
Problem Number 4
Find the indefinite integral of the function x ( ln(x) ) ^ 9.
F(x) = integral x(ln(x))^9 = (1/2)x^2 * (1/x)^9 * (1/10)x^10
Integrate by parts, letting u = (ln(x))^9 and dv = x dx so that du = 9 * 1/x * ln(x)^8 and v = x^2 / 2.
u v - int(v du) = x^2 / 2 * (ln(x))^9 - int ( x^2 / 2 * 9 / x * ln(x)^8 ) = x^2 ln(x)^9 / 2 - 9/2 int(x ln(x)^8).
This step alone would get more than half credit.
This reduces the task of integrating of x ln(x)^9 to that of integrating of x ln(x)^8. Another similar step would reduce this to an integral of x ln(x)^7, and subsequent steps would finally eliminate all integrals. Seeing that this progression exists would get an additional halfway to full credit.
Any reasonable attempt to find the resulting expression would result in full credit.
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Problem Number 5
Integrate the function function f(z) = z^ 8 + 1 / (1+( 9 z)^2 ) from z = .3 to z = 1.8.
F(z) = integrate (.3, 1.8) z^8 + 1/(1+9z^2) = z^9 / 9 + 1 / (9z^3/3) = (1/9)z^9 + x / x + (9/3)z^3 = (1/9)(1.8)^9 + 1.8 / 1.8 + (9/3)(1.8)^3 – (1/9)(.3)^9 + .3 / .3 + (9/3)(.3)^3 = .36028
1 / (9 x^3 / 3) does not have a derivative which is equal to 1 / ( 1 + (9z)^2).
To integrate this part of the expression let u = 9 z so that du = 9 dz and your integral becomse 1 / 9 * 1 / (1 + u^2) . You should recognize the antiderivative as 1/9 arcTan(u) = 1/9 arcTan ( 1 + 3 z).
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Problem Number 6
Find the general solution of dy / dt = 6 cos( 9 t) - .7.
Dy / dt = 6cos(9t) - .7 = -6sin(9t) - .7 * (9/2)t^2 * -.7t * 6t = -6sin(9t) -.7 *(9/2)t^2 * -(21/5)t
The derivative of -6sin(9t) -.7 *(9/2)t^2 * -(21/5)t is not 6 cos(9t) - .7.
An antiderivative of 6 cos(9 t) is 1/9 * 6 sin(9t) = 2/3 sin(9t).
An antiderivative of -.7 is -.7 t. So the correct antiderivative is
2/3 sin(9t) - .7 t + c.
Problem Number 7
Find the general antiderivative of sin(x) / [ 8 + cos^ 7 (x) ].
F(x) = sin(x) / 8 + cos^7x = cos(x) / 8x – sin^7x
The derivative of your result is not equal to the original function.
Let u = cos(x) so du = - sin(x) dx and your integrand sin(x) / [ 8 + cos^ 7 (x) ] dx becomes -1 / ( 8 + u^7).
I would expect this step. The resulting expression in u cannot be easily integrated at this stage in the course so I wouldn't expect more than just the substitution.
.Problem Number 8
Find the indefinite integral of the function t^ 2 cos( 6 t).
F(x) = integral t^2cos6t dx = (1/3)t^3 cos6t * -sin6t * (6/2)t^2
This requires 2 uses of integration by parts. Let u = t^2 and dv = cos(5t) dt so that du = 2 t dt and v = 1/5 sin(5 t). You get
u v - int(v du) = t^2 sin(5t) / 5 - 2 /5 int ( t sin(5 t) dt).
A similar substitution in the integral, this time letting u = t, removes reduces the task to a single simple integral.
&#Let me know if you have questions. &#