# The First Fundamental Theorem,definite integral.......

• December 29th 2006, 11:06 PM
m777
The First Fundamental Theorem,definite integral.......
Hello,
please try to solve these questions.
• December 30th 2006, 01:54 AM
CaptainBlack
Quote:

Originally Posted by m777
Hello,
please try to solve these questions.

Question #1

Show that the area under $y=x^3$ over the interval $[0,b]$ is $\frac{b^4}{4}$

This is asking for the definite integral:

$I=\int_0^b \,x^3\, dx$

You should know that $x^3$ is the derivative of $x^4/4$, and so:

$I=\int_0^b \,x^3\, dx=\left[x^4/4\right]_0^b=b^4/4-0^4/4=b^4/4$

RonL
• December 30th 2006, 02:19 AM
CaptainBlack
Question #2

(a) Evaluate the definite integral using “The First Fundamental Theorem” of calculus.

$\int_1^2 \left( \frac{1}{x^3}-\frac{2}{x^2}+\frac{1}{x^4}\right)\,dx$

This is asking you to spot what the integrand is the derivative of then by the
first fundamental theorem of calc the integral is the difference of the
derivative at the end points.

Now:

$\frac{1}{x^3}-\frac{2}{x^2}+\frac{1}{x^4} = \frac{d}{dx}\left(-\frac{1}{2x^2}+\frac{2}{x}-\frac{1}{3x^3}\right)$

So:

$\int_1^2 \left( \frac{1}{x^3}-\frac{2}{x^2}+\frac{1}{x^4}\right)\,dx = \left[-\frac{1}{2x^2}+\frac{2}{x}-\frac{1}{3x^3}\right]_1^2$ $=-\frac{1}{8}+1-\frac{1}{24}+\frac{1}{2}-2+\frac{1}{3} = -\frac{1}{3}$

RonL