Emily+and+Tara+-+Fundamental+Theorem+of+Calculus+Part+II

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=Fundamental Theorem of Calculus Part II:=

Recall:
If math F^{'}(x)=f(x) math then math \int_{a}^{b}f(x)dx=F(b)-F(a) math

Notice:
math F(b)-F(a)=\int_{a}^{b}f(x)dx math Change b to x. This changes the upper bound from a number to a variable. math F(x)-F(a)=\int_{a}^{x}f(t)dt math When you take the derivative of this equation, you get math F^{'}(x)=\frac{d}{dx}\int_{a}^{x}f(t)dt math math f(x)=\frac{d}{dx}\int_{a}^{x}f(t)dt math



Fundemental Theorem of Calculus Part II
If f is continuous on an interval I, and a is in the interval I, then for all x in the interval I,

math \frac{d}{dx}\int_{a}^{x}f(t)dt=f(x) math

Examples:
1) math \frac{d}{dx}\int_{3}^{x}ln(t^3+2t^2)dt=ln(x^3+2x^2) math 2) math \frac{d}{dx}\int_{x}^{10}e^(^t^3^+^t^)dt=-\int_{10}^{x}e^(^t^3^+^t^)=-e^(^x^3^+^x^) math 3) math \frac{d}{dx}\int_{0}^{x^2}sintdt=sin(x^2)(2x)=2xsin(x^2) math

Fundamental Theorem of Calculus Part II Proof:
Let math F(x)=\int_{a}^{x}f(t)dt math use: math F^{'}(x)=\lim_{h\to 0} \frac{F(x+h)-F(x)}{h} math notice: math F(x+h)=\int_{a}^{x+h}f(t)dt math so: math F(x+h)-F(x)=\int_{a}^{x+h}f(t)dt-\int_{a}^{x}f(t)dt=\int_{x}^{x+h}f(t)dt math

Let m be the minimum of f(t) for t from x to (x+h) and let M be the maximum of f(t) for t from x to (x+h) So: math mh\leq\int_{x}^{x+h}f(t)dt \leq Mh math

math m\leq\frac{F(x+h)-F(x)}{h} \leq M math

math {h\to 0} math

math {m, M\to f(x)} math

math {\frac{F(x-h)-F(x)}{h}\to F^{'}(x)} math

math f(x)\leq F^{'}(x) \leq f(x) math So: math F^{'}(x)=f(x) math

Show
math \frac{d}{dx}\int_{g(x)}^{h(x)}f(t)dt=f(h(x))h^{'}(x)-f(g(x))g^{'}(x) math

math \frac{d}{dx}\int_{g(x)}^{h(x)}f(t)dt= math

math \frac{d}{dx}\int_{g(x)}^{0}f(t)dt + \frac{d}{dx}\int_{0}^{h(x)}f(t)dt= math math \frac{d}{dx}\int_{0}^{h(x)}f(t)dt - \frac{d}{dx}\int_{0}^{g(x)}f(t)dt= math math f(h(x))h^{'}(x) - f(g(x))g^{'}(x) math

Proof of the Fundamental Theorem of Calculus Part I
math F(x)= \int_{a}^{x}f(t)dt math

By the Fundamental Theorem of Calculus Part II: math F^{'}=f math

Suppose math G math is any other antiderivative of math f math so math G^{'} = f math therefore math F^{'} = G^{'} math Since math \frac{d}{dx} [F - G] = 0 math math F-G=C math So math F(x)=G(x)+C math By the definition of F, we know that math F(a)=0 math So math \int_{a}^{b}f(t)dt = F(b) = F(b) - F(a) = (G(b) + C) - (G(a) + C) = G(b) - G(a) math This result shows that the definite integral math \int_{a}^{b}f(t)dt math can be evaluated using any antiderivative of f, is the first fundamental theorem of calculus. =**Technique of Integration: Subsitution**= __**Cues for Substituion**__ - Integrand contains a compostion of functions -Integrand contains a functions and its dervative

When substituting let u = inner most function

Remember the Chain Rule: math \frac{d}{dx}f(g(x))=f^{'}(g(x))g^{'}(x) math Therefore math \int f^{'}(g(x))g^{'}(x)dx=f(g(x)) math


 * Recall:**

math \int2xe^x^2dx=e^x^2+c

math let

math u=g(x), du=g^{'}(x)dx

math

math \int f^{'}(u)du=f(u)+C= f(g(x))+C math


 * Example 1:**

math \int x^3cos(x^4+2)dx math

let

math u=x^4+2, du=4x^3dx, (1/4)du=x^3dx math

math \int(1/4)cos(u)du math

math = (1/4)sin(u)+C math

math = (1/4)sin(x^4+2)+C math


 * Example 2:**

math \int(1+x^2)^1^/^2x^5dx

math

let

math u=1+x^2

x^2=u-1

x^4=(u-1)^2

math

math du=2xdx

(1/2)du=xdx

math

math =\int (u)^1^/^2(u-1)^2(1/2)du math

math =\int (u)^1^/^2(u^2-2u+1)du math

math =1/2\int (u^5^/^2-2u^3^/^2+u^1^/^2)du math

math =(1/7)u^{7/2} - (2/5)u^{5/2} + (1/3)u^{3/2} math

Substitution for Definite Integrals
- not only substitution but additionally the bounds of integration change

If math g^{'}(x) math is cont on [a,b] math f(x) math is cont on range of math g math then

math \int_{x=a}^{x=b}f(g(x))g^{'}(x)dx = \int_{u=g(a)}^{u=g(b)}f(u)du

math


 * Example 1:**

math \int_{1}^{e}(ln(x)/x) dx

math

let

math u=ln(x), du=(1/x)dx

math

if math a=x=1 u=g(a)=ln(1)=0

b=x=e u=g(b)=ln(e)=1

math

math \int_{0}^{1}udu math

math =(1/2)u^2|_{0}^{1}=1/2 math

In Class Problems:
1. math \int (x+1)(2x^2)^1^/^2dx math

2. math \int tan(x)dx math

3. math \int (ln(x)^2/x) dx math

4. math \int (1+x)/(1+x^2)dx

math

5. math \int_{0}^{1} (e^2+1)/(e^2+z)dz math

6. math \int_{-1}^{2} x(x-1)^1^/^2 math

Further Exploration:
1) math \frac{d}{dx}\int_{5}^{sinx}(t^3 - 5t^2)dt = (sin^3x - 5sin^2x)cosx = cosxsin^3x - 5cosxsin^2x math

2) math \frac{d}{dx}\int_{1}^{x^2}cos(t)dt = 2xcos(x^2) math

3) math \frac{d}{dx}\int_{1}^{arctanx}cost dt = \frac{cos(arctanx)}{1+x^2} math

4) math \int_{0}^{\frac{\pi}{2}}cosxe^{sinx}dx math

math u=sinx du=cosxdx math math sin(\frac{\pi}{2})=1 math math sin(0)=0 math

math \int_{0}^{1}e^udu = e^1-e^0=e-1 math

5) math \int_2x(x^2+4)^{100}dx math math u=x^2+4 du=2xdx math

math \int_u^{100}dx math math =\frac{u^{101}}{101} + C math math =\frac{(x^2+4)^{101}}{101} + C math