132Lecture5.pdf

132Lecture5.pdf - Lecture 5 Calculating the Volume of...

• Notes
• 6

This preview shows pages 1–2. Sign up to view the full content.

Lecture 5: Calculating the Volume of Solids 1 A useful formula about substitution rule Before starting the lecture we mention a very useful formula that saves you time when doing substitution rule. If f is continuous on [ a, b ] and F is one of its antiderivative, then for any constant k 6 = 0, Z f ( kx ) dx = F ( kx ) k + C ; (1) and Z b a f ( kx ) dx = F ( kx ) k | b a = F ( kb ) k - F ( ka ) k . (2) Using the formula, it is immediate to see that 1. R cos( kx ) dx = sin( kx ) k + C, R sin( kx ) dx = - cos( kx ) k + C, 2. R sec 2 ( kx ) dx = tan( kx ) k + C, R sec( kx ) tan( kx ) dx = sec( kx ) k + C, 3. R 1 1 - ( kx ) 2 dx = arcsin( kx ) k + C, R 1 1+( kx ) 2 dx = arctan( kx ) k + C, 4. R e kx dx = e kx k + C . The second thing we want to mention is when doing the WebAssign on this section and previous one, you can evaluate the definite integral at . For example, if you are going to evaluate R π 24 0 (6 cos(4 x ) - 6 sin(8 x )) dx , you can type the following integrate 6 cos(4x) - 6 sin(8x) from 0 to pi/24 . 2 Volume formula of solids We firstly derive the volume formula for solids. The idea is similar to the Riemann sum used to define the definite integral: Divide a solid into n slices with equal width Δ x by planes perpendicularly intersecting x -axis at points x 0 = a, x 1 , x 2 , · · · , x n = b .

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.
• Summer '08
• WOLBACH

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern