Question

Can you explain definite integrals?

Answer 1

Integrals are often used to determine the area under a curve/surface depending on the number of integrals invovled. Assuming that just one integral is being used, then
\[\int\limits_{ }^{ }f(x)dx\]
is the area between the curve f(x) and the x axis.
When we have a definite integral, values are given which allow the area under the curve to be calculated between two specific points:
\[\int\limits_{a }^{ b}f(x)dx\]
This gives the total area under the curve starting at x = a and ending at x=b.

Answer 2

id agree in principal, but the first part is alittle askew

Answer 3

the first part provides us with a family of related functions that can be used in a variety of ways

Answer 4

\[\int f(x)dx;\ knowing\ that\ F(0)=5\]

Answer 5

How do I solve this problem?? \[\int\limits_{-1}^{5}(1+3x)dx\]

Answer 6

integrate it as usual; to find the desired function; then evaluate that function using the limits as: F(5)-F(-1)

Answer 7

How do I integrate it?

Answer 8

do you know how to find a derivative?

Answer 9

yes

Answer 10

integration is just undoing a derivative; also known in some earlier circles as antidifferentation

Answer 11

\[\frac{d}{dx}? = Constant\]

\[? =\int Constant\]

Answer 12

\[\frac{d}{dx}\frac{x^{n+1}}{n+1}=x^{n}\]

\[\frac{x^{n+1}}{n+1}=\int x^{n}\]

Answer 13

so I'm supposed to do that with 1+3x? like the integral would be x+\[3/2x ^{^{2}}\]

Answer 14

exactly

Answer 15

\[F(x)=x+\frac{3}{2}x^2\]
to find the definite part of it; use the limits
F(2)- F(-1)

Answer 16

\[F(2)=2+\frac{3}{2}(2)^2\]\[-F(-1)=-(-1+\frac{3}{2}(-1)^2)\]

Answer 17

so now I have 5.5. Is that right? and if so, what do I do next?

Answer 18

2+6+1-3/2 = 9 - 3/2 = 8 - 1/2 = 7.5

Answer 19

what you do next is put 7.5 in the answer box and move to the next problem ;)

Answer 20

How did you get 7.5? and the back of the book says that the answer is 42...

Answer 21

i plugged in the limits, and did the math; as typed out above

Answer 22

and 42 is not the answer unless there is more to the problem then you have posted

Answer 23

i see a mistake i did; is read 5 as a 2

Answer 24

So how do I fix that? What do I do differently?

Answer 25

I put in 5 instead of 2, and I got 43, not 42...

Answer 26

\[F(5)-F(-1)=5+\frac{3}{2}(5)^2-(-1+\frac{3}{2}(-1)^2)\]\[\hspace{10em} =5+\frac{75}{2}+1-\frac{3}{2}\]

Answer 27

6 + 72/2 = 6 + 36 = 42

Answer 28

ooh yay it's right!! Thanks so much!!

Answer 29

yep, yw ;)

Answer 30

How can I solve the problem using the following formula?
\[\int\limits_{a}^{b}f(x)dx=\lim_{n \rightarrow \infty} \sum_{i=1}^{n}f(x _{i})\Delta x\]