Fundamental Theorem of Calculus

**c_323_h** · June 27th 2006, 05:10 PM

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.

**malaygoel** · June 27th 2006, 06:56 PM

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

**c_323_h** · June 28th 2006, 06:43 AM

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

**ThePerfectHacker** · June 28th 2006, 07:37 AM

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

**topsquark** · June 28th 2006, 08:39 AM

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,
$<br /> y'=-F'(1-3x)(-3)=3F'(1-3x)$
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!

-Dan

**ThePerfectHacker** · June 28th 2006, 10:46 AM

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!

-Dan

I know, i am the best.

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