The Indefinite Integral
Integration We begin with a question. Question: List two functions F(x) such that F'(x) = x Answer: 1/2 x^{2} and 1/2 x^{2} + 3 We can see that if F'(x) = x then F(x) = 1/2 x^{2} + C for some constant C. We call F(x) the antiderivative or integral of f(x) and write
In general, if F'(x) = f(x) then we write
From the derivative formula
d We get the integral formula
Just like with derivatives, to find an antiderivative of a sum or difference, we can take the antiderivative of each term. Also like derivatives, the antiderivative of the product or quotient is not easily found.
Example Which of these has an easy to find antiderivative A. 8x^{3} - 6x B.
x
Solution We can find the antiderivative of part A. easily, by finding the antiderivative of 8x^{3} and 6x separately. The antiderivative is 8(1/4 x^{4}) - 6(1/2 x^{2}) + C = 2x^{4} - 3x^{2} + C B. This one, on the other hand, is a quotient. We do not have a way of finding its antiderivative.
Exercise Find the following integrals:
Particular Solutions We have seen that an integral produces a whole family of solutions parameterized by C. In most applications, we are given an initial or other condition and hence find the value of C. The antiderivative with known C is called a particular solution.
Example Find a solution to F'(x) = 4x - 3 given that F(1) = 2
Solution: We first find an antiderivative: F(x) = 2x^{2} - 3x + C Now plug in 1 for x and 2 for F to get: 2 = 2(1)^{2} - 3(1) + C = -1 + C So that C = 3. The particular solution is F(x) = 2x^{2} - 3x + 3.
Example
Applications Since the acceleration of gravity is a constant a = 32, we can derive the physics equations.
Example Suppose that we kick a football with an initial upward velocity of 100 feet per second how long will it take to hit the ground?
Solution We have v(t) = -32 dt = -32t + C v(0) = 100 = C s(t) = (-32t + 100)dt = -16t^{2} + 100t + C s(0) = 0 = C hence s(t) = -16t^{2} + 100t = t(-16t + 100) So that s(t) = 0 when -16t + 100 = 0 or t = 100/16 = 6.25 It will take 6.25 seconds to hit the ground.
Example Suppose the marginal revenue for a ski resort is M = 50 - 0.01 x And suppose that at $50 per ticket, the ski resort will have 2,000 skiers. Find the demand equation.
Solution Since the marginal revenue is the derivative of the revenue, the revenue is the antiderivative of the marginal revenue. R = (50 - 0.01x)dx = 50x - 0.005 x^{2} + C The revenue is equal to the price times the quantity. That is 50x - 0.005 x^{2} + C = px Now find C by noting that when p = 50, x = 2,000. 50(2,000) - 0.0005 (2,000)^{2} + C = (50)(2,000) 80,000 + C = 1,000,000 C = 920,000 Substituting the C into our equation and dividing by x gives the demand equation p = 50 - 0.005 x + 920,000/x
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