# Fundamental Theorem of Calculus

• June 27th 2006, 05:10 PM
c_323_h
Fundamental Theorem of Calculus
Need some help with FTC problems. My calculus textbook defines the FTC in two parts:

Definition Part 1:
If $f$ is a continuous function on $[a,b]$, then the function $g$ defined by $g(x) = \int_a^x f(t)dt$ where $a\leq x\leq b$
is continous on $[a.b]$ and differentiable on $(a,b)$, and $g'(x) = f(x)$

Definition Part 2:
If $f$ is a continuous function on $[a,b]$, then $\int_a^b f(x)dx=F(b)-F(a)$ where F is any antiderivative of $f$, that is, a function such that $F'=f$

Whew, here are the problems I'm confused on...

Problem 1. Using Part 1 definition, find the derivative of the function:
$y=\int_{1-3x}^1 \frac{u^3}{1+u^2}du$

Problem 2. Using Part 2 defintion, evaluate the integral:
$\int_0^2 x(2+x^5)dx$

Thanks a lot.
• June 27th 2006, 06:56 PM
malaygoel
Quote:

Originally Posted by c_323_h
Need some help with FTC problems. My calculus textbook defines the FTC in two parts:

Definition Part 1:
If $f$ is a continuous function on $[a,b]$, then the function $g$ defined by $g(x) = \int_a^x f(t)dt$ where $a\leq x\leq b$
is continous on $[a.b]$ and differentiable on $(a,b)$, and $g'(x) = f(x)$

Definition Part 2:
If $f$ is a continuous function on $[a,b]$, then $\int_a^b f(x)dx=F(b)-F(a)$ where F is any antiderivative of $f$, that is, a function such that $F'=f$

Whew, here are the problems I'm confused on...

Problem 1. Using Part 1 definition, find the derivative of the function:
$y=\int_{1-3x}^1 \frac{u^3}{1+u^2}du$

Problem 2. Using Part 2 defintion, evaluate the integral:
$\int_0^2 x(2+x^5)dx$

Thanks a lot.

Problem 1
g(x)=y
f(x)= $\frac{u^3}{1+u^2}$
Problem 2
Intregrate the integrand and put the limits

Keep Smiling
Malay
• June 28th 2006, 06:43 AM
c_323_h
Quote:

Originally Posted by malaygoel
Problem 1
g(x)=y
f(x)= $\frac{u^3}{1+u^2}$
Problem 2
Intregrate the integrand and put the limits

Keep Smiling
Malay

hmmm....i don't know how to integrate the integrand. that is why i am asking
• June 28th 2006, 07:37 AM
ThePerfectHacker
Quote:

Originally Posted by c_323_h

Problem 1. Using Part 1 definition, find the derivative of the function:
$y=\int_{1-3x}^1 \frac{u^3}{1+u^2}du$

That is same as,
$y=-\int_1^{1-3x}\frac{u^3}{1+u^2}du$
Consider a function,
$F(x)=\int_1^x\frac{u^3}{1+u^2}du$
by, FTC we have,
$F'(x)=\frac{x^3}{1+x^2}$
Now note that,
$y=-(F\circ 1-3x)(x)=-F(1-3x)$
A function composition!
Then by the Chain Rule,
$
y'=-F'(1-3x)(-3)=3F'(1-3x)$

But,
$F'=\frac{x^3}{1+x^2}$
Thus,
$y'=\frac{3(1-3x)^3}{1+(1-3x)^2}$
• June 28th 2006, 08:39 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
That is same as,
$y=-\int_1^{1-3x}\frac{u^3}{1+u^2}du$
Consider a function,
$F(x)=\int_1^x\frac{u^3}{1+u^2}du$
by, FTC we have,
$F'(x)=\frac{x^3}{1+x^2}$
Now note that,
$y=-(F\circ 1-3x)(x)=-F(1-3x)$
A function composition!
Then by the Chain Rule,
$
y'=-F'(1-3x)(-3)=3F'(1-3x)$

But,
$F'=\frac{x^3}{1+x^2}$
Thus,
$y'=\frac{3(1-3x)^3}{1+(1-3x)^2}$

Can I just butt in and say that I've never seen anything like that and Dude! That was TOTALLY awesome! :cool:

-Dan
• June 28th 2006, 10:46 AM
ThePerfectHacker
Quote:

Originally Posted by topsquark
Can I just butt in and say that I've never seen anything like that and Dude! That was TOTALLY awesome! :cool:

-Dan

I know, i am the best.