When integrating, are you finding.. (2 Viewers)

[sinophile]

The area below the curve, i.e the "scalar quantity between the curve and the x-axis"?

or are you just finding a "real number", which could either be positive or negative?

 

[MC Squidge]

an amount

 

[sinophile]

an amount

So, a number or an area, which is a scalar quantity?

 

[samthebear]

i thought it was either depending on the question =/

 

[Dumbledore]

a number, this concept is usually asked in a question like
"explain why integral(0, 2) x^2-1 dx will not find the area of this curve bounded by the x axis

 

[hermand]

when asked to integrate, ie
you will get zero as an answer, because the negative area [below the x axis] will cancel out the positive area [above the axis], because when you are asked to integrate, you can get negative numbers,

but when asked to find the area, you must absolute value everything because you can't have a negative area.

 

[undalay]

Scalar quantities can be negative.

 

[kwabon]

The area below the curve, i.e the "scalar quantity between the curve and the x-axis"?

or are you just finding a "real number", which could either be positive or negative?

an area cannot be negative ... so hence ... u are finding an amount.

 

[Trebla]

When you find a DEFINITE INTEGRAL:

...you are finding the summation of infinitely small partitions (think of them as tiny rectangle approximations) of a region bounded by the curve y = f(x), x = a and x = b (assuming the integral and function is well defined across that x interval). Each partition has an infinitely small width dx and length f(x). Hence when f(x) is negative then f(x) dx is negative, hence it contributes negatively to the integral summation.

 

[Gibbatron]

The area below a curve has to be positive because it is an area. If you are asked to find an area between two points, and the line cuts the x axis somewhere in betwen these two points, you will need to integrate twice, once for the area under the x axis and once for above, find the modulus of both areas and add them. It is impossible to get a negative area.