Antiderivatives and Differential Equations |
Doing Differential Calculus is like zooming in on tiny details with the camera lens. We spend our time discussing the points that make up the line of a given function's curve -- the slopes of the tangents at particular points, the instantaneous velocity and acceleration at given points in time, etc.
With Integral Calculus, we pan back with the camera, we move into the third dimension and consider not the points that make up the actual line of the curve -- now we find areas under curves and volumes of solids created by revolving areas under curves around given axes.
This explains why, when we differentiate, we lower the power of the variable by one dimension -- but when we antidifferentiate, we raise the power of the variable by one dimension.
So, in reality, an antiderivative is the function we differentiated to get the expression we have to antidifferentiate. We're going backwards.
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Antiderivatives -- the power rule
The power rule for differentiation says if f (x) = ax^{ n} then f '(x) = anx^{ n – 1} .
Obviously then, since we must go backwards and undo what we did, we're going to have to raise the power of the variable by 1 -- and then we must adjust the coefficient so that we get exactly
f (x) and not some multiple of it.
There's one other issue to consider -- the fact that the derivative of a constant = 0. Since any constant that appeared in f (x) will not appear in f '(x) we must indicate in our antiderivative that there might have been a constant in the function.
Notation: If we're asked to find an antiderivative of an expression labeled f (x) rather than f '(x) , we label our antiderivative F(x). If the derivative is labeled f '(x), the antiderivative will be labeled f (x).
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Example 1
Find the most general antiderivative of f (x) = 5x^{ 3}
Solution
Since the power of x in the derivative = 3, the original function must have been of degree 4.
So we know that F(x) = ___ x^{ 4} + C. When we differentiate F(x), we're going to multiply the coefficient ___ by 4, therefore ___ has to be 5/4 because we need f (x) to = 5x^{ 3}.
If f (x) = ax^{ n} , then the most general antiderivative of f is: where C is an arbitrary constant and n is not = –1. |
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Notice that the new exponent becomes the denominator. We have to divide a by n + 1, so the way to proceed is to first establish the value of n + 1, write the variable to that power and then figure out the coefficient like in Example 1. Once we realized that n + 1 was 4, we had to divide the coefficient a = 5 by 4. That's how we got for our antiderivative.
To check, we simply differentiate F(x) to get f (x).
In the case of negative exponents, we must be careful to raise powers so if we are antidifferentiating x^{ – 4}, we will get __ x^{ – 3}.
In the case of fractional exponents, when we add 1 to the exponent, we think of it as so that we have our common denominator immediately.
For instance -- if f (x) = __ x^{ }1/3, then F(x) = __ x^{ 1/3 + 3/3} = __ x^{ 4/3} + C.
Had the exponent been 4/7, we would think of 1 as 7/7.
**Special Case:
When n = – 1, the power rule doesn't apply. Recall that the derivative of ln x is which of course is x^{ – 1}, so the antiderivative of ax^{ – 1} is a ln x + C.
Find the most general antiderivative for:
a) f (x) = 5x^{ 3} – 7x^{2} + 3x – 9 | b) | c)** |
**In question (c), rewrite and rewrite .
When given a fraction such as , simply divide each term in the numerator by the denominator (x^{ 1/ 2 }) and then antidifferentiate.
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the power rule | examples1 | differential equations |
examples2 | practice | solutions |
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When solving a differential equation, we'll be given a derivative at some level -- be it first, 2nd or some other derivative and we'll be given boundary values -- information that enables us to find the exact value of C at each level.(note: use different letters for the constant a different levels of antidifferentiation.)
Example 1: Find f (x) if f '(x) = 2x + 3 and f (0) = –5
Solution: Using the power rule we know that f (x) = x^{ 2} + 3x + C
From the boundary value we know that when x = 0, f (x) = – 5 so
–5 = 0 + 0 + C, therefore C = –5
so f (x) = x^{ 2} + 3x – 5
Example 2: Find f (x) if f "(x) = 2x + 3 with f '(0) = –5, and f (0) = 3
Solution: Using the power rule we know that f '(x) = x^{ 2} + 3x + C
From the boundary value we know that when x = 0, f '(x) = – 5
so f '(x) = x^{ 2} + 3x – 5
Which means that
Since f (0) = 3, we know D = 3
So
Note: there are questions where we're given 2 boundary values at the same level -- the function level. In such a case, we just set up and solve a system of 2 equations in 2 unknowns.(see ex.1)
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the power rule | examples1 | differential equations |
examples2 | practice | solutions |
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1/ Find f (x) if f "(x) = 6x + 6, with boundary values f (0) = 4 and f (1) = 3
f '(x) = 3x^{ 2} + 6x + C by antidifferentiation so f (x) = x^{ 3} + 3x^{ 2} + Cx + D
We know that f (0) = 4 therefore 4 = D
So, f (x) = x^{ 3} + 3x^{ 2} + Cx + 4, and f (1) = 3
Substituting x = 1 and f (1) = 3, we get 3 = 1 + 3 + C + 4,
Therefore C = –5
Which means that f (x) = x^{ 3} + 3x^{ 2} – 5x + 4
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2/ Find f (x) if , f (1) = 1 and f '(1) = 2.
By antidifferentiation,
Since f '(1) = 1, we substitute to find C
So,
Which means that
So now we can say that
And since f (1) = 1, we substitute to find D.
Therefore
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3/ Find f (x) if f '(x) = 3 cos x + 5 sin x and f (0) = 4
By antidifferentiation, f (x) = 3 sin x – 5 cos x + C
Substituting x = 0 and f (0) = 4 we get 4 = 3 sin(0) – 5 cos(0) + C
Therefore, since sin(0) = 0 and cos(0) = 1, C = 9
So, f (x) = 3 sin x – 5 cos x + 9
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the power rule | examples1 | differential equations |
examples2 | practice | solutions |
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1/ Find the most general antiderivative of the functions listed:
a) f (x) = 9x^{2} – 4x + 3 | b) f (x) = 10x^{ 4} – 6x^{ 3} + 5 | c) |
d) | e) | f) (hint: divide) |
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2/ Solve these differential equations subject to the given conditions:
a) f '(x) = 12x^{ 2} – 6x + 1, f (1) = 5 | b) f "(x) = 4x + 1, f '(2) = –2 and f (1) = 3 |
c) f '"(x) = 6x, f "(0) = 2, f '(0) = –1, f (0) = 4 | d) f "(x) = x + cos x, f '(0) = 2, f (0) = 1 |
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3/ A point moves on a coordinate line with a(t) = 2 – 6t.
If the initial conditions are v(0) = –5, and s(0) = 4, find s(t).
(reminder: if s(t) is the position function, v(t) = s '(t) and a(t) = s "(t).)
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4/ The volume V of a balloon is changing with respect to time t at a rate of
. If at t = 4, the volume = 20 cm^{ 3} , express V as a function of t.
5/ The rate of change of the temperature T of a solution (in ^{o }C) is with t in minutes.
If T = 5 at t = 0, find a formula for T at time t.
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the power rule | examples1 | differential equations |
examples2 | practice | solutions |
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1/ Find the most general antiderivative of the functions listed:
a) f (x) = 9x^{2} – 4x + 3 | b) f (x) = 10x^{ 4} – 6x^{ 3} + 5 | c) |
F(x) = 3x^{ 3} – 2x^{ 2} + 3x + C | F(x) = 2x^{ 5} – x^{ 4} + 5x + C | |
d) | e) | f) |
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2/ Solve these differential equations subject to the given conditions:
a) f '(x) = 12x^{ 2} – 6x + 1, f (1) = 5 | b) f "(x) = 4x + 1, f '(2) = –2 and f (1) = 3 |
f (x) = 4x^{ 3} – 3x^{ 2} + x + C f (1) = 5 = 4 – 3 + 1 + C, so C = 3 f (x) = 4x^{ 3} – 3x^{ 2} + x + 3 |
f '(x) = 2x^{ 2} + x + C, f '(2) = –2 so C = –12 , f (1) = 3, so |
c) f '"(x) = 6x, f "(0) = 2, f '(0) = –1, f (0) = 4 | d) f "(x) = x + cos x, f '(0) = 2, f(0) = 1 |
f "(x) = 3x^{ 2} + C, with C = 2 f '(x) = x^{ 3} + 2x + D with D = –1 , with E = 4 so, |
f '(x) = + sin x + C, with C = 2 f (x) = – cos x + 2x + D, with D = 2 so, f (x) = – cos x + 2x + 2 |
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3/ A point moves on a coordinate line with a(t) = 2 – 6t.
If the initial conditions are v(0) = –5, and s(0) = 4, find s(t).
Solution: since s"(t) = 2 – 6t, s '(t) = v(t) = 2t – 3t^{ 2} + C, but v(0) = –5
so, v(t) = 2t – 3t^{ 2} – 5, and therefore s(t) = t^{ 2} – t^{ 3} – 5t + D
since s(0) = 4, D = 4 so s(t) = t^{ 2} – t^{ 3} – 5t + 4.
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4/ The volume V of a balloon is changing with respect to time t at a rate of
. If at t = 4, the volume = 20 cm^{ 3} , express V as a function of t.
solution: we antidifferentiate dV/dt to get V as a function of t.
so, , with V(4) = 20
therefore, C = 2, so .
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5/ The rate of change of the temperature T of a solution (in ^{o }C) is , t in minutes.
If T = 5 at t = 0, find a formula for T at time t.
solution: we antidifferentiate dT/dt to get T as a function of t.
The antiderivative of dT/dt is T = , and we know that T(0) = 5, so C = 5.
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the power rule | examples1 | differential equations |
examples 2 | practice | solutions |
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