Hello,

please try to solve these questions.

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- Dec 29th 2006, 11:06 PMm777The First Fundamental Theorem,definite integral.......
Hello,

please try to solve these questions. - Dec 30th 2006, 01:54 AMCaptainBlack
Question #1

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

This is asking for the definite integral:

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

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

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

RonL - Dec 30th 2006, 02:19 AMCaptainBlack
Question #2

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

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

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:

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

So:

$\displaystyle \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$$\displaystyle =-\frac{1}{8}+1-\frac{1}{24}+\frac{1}{2}-2+\frac{1}{3} = -\frac{1}{3}$

RonL