Suppose f integrable (Note f continuous implies f integrable).

If n equal subdivisions: Δx = \frac{b−a}{n} and if we use righthand endpoints: x_{i} = a + iΔx = a + \frac{(b−a)i}{n}

∫_{a}^{b} f(x)dx = lim_{n→∞}Σ_{i=1}^{n}f(a + \frac{(b−a)i}{n})(\frac{b−a}{n})

Evaluate the limit by recognizing the sum as a Riemann sum for a function defined on [0, 1]

1.) lim_{n→∞}Σ_{i=1}^{n}sin(\frac{i}{n})\frac{1}{n}

2.) lim_{n→∞}Σ_{i=1}^{n} \frac{i^{5}}{n^{6}}

The Fundamental Theorem of Calculus: Suppose f continuous on [a, b].

1.) If G(x) = ∫_{a}^{x} f(t)dt, then G′(x) = f(x).

2.) ∫_{a}^{b} f(t)dt = F (b)−F (a) where F is any antiderivative of f, that is F′ = f.

The Fundamental Theorem of Calculus: Suppose f continuous on [a, b]. 1.) If \frac{d}{dx}[∫_{a}^{x} f(t)dt] = f(x). 2.) ∫_{a}^{b} F′(t)dt = F (b) − F (a).

Examples: 1.) If G_{1}(x) = ∫_{0}^{x} t^{2}dt, then G′_{1}(x) = . 2.) If G_{2}(x) = ∫_{5}^{x} t^{2}dt, then G′_{2}(x) = . 3.) If G_{3}(x) = ∫_{−2}^{x} sin(t^{2})dt, then G′_{3}(x) = . 4.) If G_{4}(x) = ∫_{4}^{x} tan(\frac{t^{3}}{t+1})dt, then G′_{4}(x) = . 5.) If G_{5}(x) = ∫_{1}^{x} \sqrt{3t − 5}dt, then G′_{5}(x) = . Examples: 1.) If H_{1}(x) = ∫_{0}^{x^{3}} t^{2}dt, then H′_{1}(x) = . 2.) If H_{2}(x) = ∫_{5}^{x^{3}} t^{2}dt, then H′_{2}(x) = . 3.) If G(x) = ∫_{2}^{ln(x)} \frac{t}{t+1}dt, then G′(x) = .

4.) If G(x) = ∫3^{\frac{1}{x}} sec(t)dt, then G′(x) = .

The Fundamental Theorem of Calculus: Suppose f continuous on [a, b]. 1.) If G(x) = ∫_{a}^{x} f(t)dt, then G′(x) = f(x). Proof G′(x) = lim_{h→0} \frac{G(x+h)−G(x)}{h},

= lim_{h→0}^{\frac{∫_{a}^{x+h} f(t)dt−∫_{a}^{x} f(t)dt}{h}}

= lim_{h→0}^{\frac{∫_{x}^{x+h} f(t)dt}{h}}

≤ lim_{h→0}^{\frac{∫_{x}^{x+h} M_{h}dt}{h}} where M_{h} = max{f(t) | x ≤ t ≤ t + h} (Note M_{h} exists by extreme value thm)

≤ lim_{h→0} \frac{(M_{h})(h)}{h} ≤ lim_{h→0}M_{h} = f(x) Similarly G′(x) ≥ f(x) (using m_{h} = min{f(t) | x ≤ t ≤ t + h} The Fundamental Theorem of Calculus: Suppose f continuous on [a, b]. 2.) ∫_{a}^{b} f(t)dt = F (b)−F (a) where F is any antiderivative of f, that is F′ = f. Proof Let G(x) = ∫_{a}^{x} f(t)dt. Then G′(x) = f(x) (ie, G is an antiderivative of f). Let F be any antiderivative of f. Then F (x) = G(x) + C = ∫_{a}^{x} f(t)dt + C for some constant C. Thus F (b) − F (a) = G(b) + C − [G(a) + C] = G(b) − G(a) = ∫_{a}^{b} f(t)dt − ∫_{a}^{a} f(t)dt = ∫_{a}^{b} f(t)dt.

∫_{0}^{π} cos^{3}(x)dx = ∫_{0}^{π} cos^{2}(x)cos(x)dx = ∫_{0}^{π}(1−sin^{2}(x))cos(x)dx

= ∫_{0}^{0}(1 − u^{2})du = 0

Let u = sin(x), du = cos(x)dx, when x = 0, u = sin(0) = 0, when x = π, u = sin(π) = 0 Shortcut method: Use symmetry. For example: If f is an odd function (f(−x) = −f(x)), then ∫_{−a}^{a} f(x)dx = 0 If f is an even function, then ∫_{−a}^{a} f(x)dx = 2 ∫_{0}^{a} f(x)dx