INTEGRATION: THE FUNDAMENTAL THEOREM OF CALCULUS MR. VELAZQUEZ AP CALCULUS
RECALL: ANTIDERIVATIVES When we last spoke of integration, we examined a physics problem where we saw that the area under the velocity curve gave us the distance traveled by the object; it was also implied that distance could be considered the antiderivative of the velocity curve. Today we will be drawing attention to this connection between antiderivatives and definite integrals. Remember that for a continuous and differentiable function f(x), we use F(x) to represent the general antiderivative of f over a given interval.
FUNDAMENTAL THEOREM OF CALCULUS, PT. 1 Let f(x) be continuous and differentiable on a, b and let g(x) be defined as follows: x g x = න a f t dt Then g(x) is continuous and differentiable on a, b and g t = f t Or, using alternate notation: x d dx න f t dt = f x a In other words, taking the derivative of an integral where the variable of differentiation is in the limit of the integral will result in simply the integrand.
x Differentiate g x = 4 e 2t cos 2 1 5t dt. There s not much to this one. Since the lower limit of the integral is a number and the upper limit is simply x, all we need to do is apply part one of the FTC. This simply means we take the integrand, replace t with x, and that will be our answer. x g x = d dx න e 2t cos 2 1 5t dt 4 g x = e 2x cos 2 1 5x BE CAREFUL! Our job is only this easy if we can verify that (a) the lower limit of the integral is a number, and (b) the upper limit is just x (if it s a function of x, we d have to approach it differently).
1 t Differentiate h x = 4 +1 dt. x 2 t 2 +1 The first thing we notice is that the FTC requires the lower limit to be a number and the upper limit to be the variable. We can fix this using a property of definite integrals: d 1 dx න t 4 + 1 x 2 t 2 + 1 dt = d x 2 dx න t 4 + 1 1 t 2 + 1 dt = d x 2 dx න t 4 + 1 1 t 2 + 1 dt Then we notice that FTC requires x in the upper limit, but what we have is x 2. Your instinct might tell you just substitute x 2 instead of x which is true, but since we re taking a derivative, we cannot forget chain rule (let u = x 2 ): d x 2 dx න t 4 + 1 1 t 2 + 1 x d 2 dt = du න t 4 + 1 1 t 2 + 1 dt = 2x x8 + 1 x 4 + 1 du dx = x2 4 + 1 x 2 2 + 1 2x
A USEFUL FORMULA FOR FTC, PT. 1 If the upper and lower limits of the integral are both functions of x which we ll call v(x) and u(x) (i.e. neither one is a constant), then we can use the formula below to differentiate these integrals: d u(x) dx න v(x) f t dt = v x f v x + u x f u x Example: Differentiate න 3x x t 2 sin 1 + t 2 dt v x = x v x = 1 2 x u x = 3x u x = 3 Applying the formula above: 3x d dx න t 2 sin 1 + t 2 dt = 1 x 2 sin 1 + x 2 + 3 3x 2 sin 1 + 3x 2 x 2 x = 1 2 x sin 1 + x + 27x2 sin 1 + 9x 2
Find d dx x 1 t cos t 2 dt. Find d 1 (t 2 + 1) dt. dx 2x+1
Find d dx 3x 2 e t dt. 2x
FUNDAMENTAL THEOREM OF CALCULUS, PT. 2 Let f(x) be continuous and differentiable on a, b and let F x be any antiderivative of f. Then: b b න f x dx = F x ቚ = F b F(a) a a In other words, to compute a definite integral, we can take the antiderivative of the integrand, evaluate it for b and a, and then subtract the latter from the former. Note that we can use any antiderivative, since the constant terms will always cancel each other out.
EXAMPLE: Here is an example we saw in our previous lecture: 0 2 x 2 + 1 dx If we look at the integrand f x = x 2 + 1, we can easily identify an antiderivative: F x = 1 3 x3 + x + C We can ignore C, since it will always cancel out when we do the subtraction. The integral can therefore be computed the following way: 2 න x 2 + 1 dx = 1 0 3 x2 + x 2 0 = 1 3 2 3 + 2 1 3 0 3 + 0 = 8 3 + 6 3 + 0 = 14 3 (Much easier than using a Riemann sum, isn t it?)
2 න t 2 + t 2 dt = 1 Evaluate the following definite integrals: න 2 1 t 2 + t 2 dt =
2π න 3 sin x dx = 0 Evaluate the following definite integrals: ln 3 න 2e x dx = 0
Evaluate the following definite integrals: න 0 π 2 cos x dx = e x 2 + 1 න dx = 1 x
x + 1 x 0 Suppose f x = ቊ sec 2 x x > 0 π 4 Find න f(x) dx 1
CLASSWORK: FUNDAMENTAL THEOREM OF CALCULUS 1. 2. 3. MATH JOURNALS WILL BE CHECKED TODAY! For classwork, complete the four problems shown to the left. HW: Khan Academy (Due 12/14)