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Antiderivative & Indefinite Integrals

Antiderivative & Indefinite Integrals. Chapter 3.9 pgs. 119 – 122 Brandon Kyle. You can actually use this stuff.

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Antiderivative & Indefinite Integrals

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  1. Antiderivative & Indefinite Integrals Chapter 3.9 pgs. 119 – 122 Brandon Kyle

  2. You can actually use this stuff • Antiderivatives can be used in the real world when given the acceleration of an object (such as an object in freefall will have the acceleration of gravity, -9.81 m/s2) you can find the velocity, or if given the velocity, you can find the distance traveled.

  3. You will learn how to find the original function, when given the derivative of a function. If f’(x) or f-prime is given, you can find f(x).

  4. The past • Earlier in chapter 3, you learned how to find the derivative of a function when given a function. Now you will learn how to find a function when given the derivative of the function. This is called the antiderivative. • Some tricky parts in this section will be the antiderivatives of trig. functions. Keep in mind the properties of section 3.6 which says that deriv of sine = cosine, however the deriv of cosine = -sine, and also that you are working backwards

  5. Typical Problem • Given: f’(x) = x4 find f(x) = ? • In general, do the opposite of taking the derivative • Derivative: nxn-1 Antiderivative: (xn+1)/n +1 • Therefore f(x) = (1/5)x5 + C • Always add C because deriv of C = 0

  6. Trig. problem • Given: f’(x) = sin(x), find f(x) = ? • If cos(x) d/dx = -sin(x), then • –cos(x) d/dx = sin(x) • So f(x) = -cos(x)

  7. Trig. Problem cont. • Given: f’(x) = sin(5x), find f(x) = ? • d/dx(5x) * ? = 1 • ? = 1/5 • d/dx ? = sin, ? = -cos • Keep inside function(5x) • so: f(x) = -1/5cos5x

  8. Step by step, day by day • Add 1 to exponent (f’(x) = x4, so 4 + 1 = 5) • Divide problem by n + 1(x5 / 5 or (1/5)x5) • Write + C to signify that a constant could have been included(f(x) = (1/5)x5 + C)

  9. Sky diving w/o a parachute (oh nooo) Given: v(t) = -9.81t, where v(t) is meters per second, and t is time. • Find distance fallen after 2 sec. • Antiderivative of velocity is position function, v(t) = p’(t) • p(t) = ?

  10. Sky diving cont. • p(t) = -4.9t2 + C • p(2) = -4.9(2)2 + C = -19.6 + C • position at 2 sec = -19.6 m/s

  11. Now you know… • Now you should know how to find a function or antiderivative when given the derivative of the function.

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