Question

How to find total area of the region between curve and x-axis???
y=2x+7; 1≤x≤5

and

y=(4x)/(1+x^2), interval -1≤x≤1

Answer 1

For f(x)=2x+7, and \(as\ long\ as\ f(x)\ge0\ in\ the\ given\ interval\ [1,5]\), the area is given by
\(\int_1^5 f(x)dx\).

Answer 2

what about the secondn one?

Answer 3

@kevst3r
Can you tell me if f(x)=(4x)/(1+x^2) is even or odd?

Answer 4

odd

Answer 5

For an odd function, f(-x)=-f(x).
IF f(x) in the interval [0,1] is >= 0, then f(x) in the interval [-1,0] is <=0.
Thus the total area between the curve and the x-axis is equal to
\(\int_0^1 f(x)dx - \int_{-1}^0 f(x)dx\), or simply using the properties of odd functions:
\(\int_0^1 f(x)dx - \int_{-1}^0 f(x)dx = 2\int_0^1 f(x)dx\)

Answer 6

the answer is \[(1/3)\ln(lnx)+C\]

Answer 7

@mathmate so how do ig et that

Answer 8

Integrate f(x)=4x/(1+x^2)

Note that (except for constants), the numerator is the derivative of the denominator.
Use then substitution s=(1+x^2), ds=2x dx, or 2ds = 4x dx
Make the appropriate substitutions,
\(\int f(x)dx = \int \frac{4xdx}{1+x62}=\int 2ds/s = 2log(s)=2 log(1+x^2)+C\)
but C disappears (cancels out) when you evaluate definite integrals:
\(2\int_0^1 f(x)dx = 2\int_0^1 \frac{4xdx}{1+x^2}dx=2[log(1+x^2)]_0^1=2*log(2)=4log(2)\)
where log(x) is natural log (usually written as ln(x) in highschool).

Answer 9

Typo:
\(2\int_0^1 f(x)dx = 2\int_0^1 \frac{4xdx}{1+x^2}dx=2[2log(1+x^2)]_0^1=2*2log(2)=4log(2)\)