Use define integrals to calculate the area under the graph of y=tanx between x=a to x=b. (hint: write tangent in the form of sinx/cosx)
How would you do this?
area=int (sinx/cosx)dx
let Y=cosx
dY=-sinx dx
area=int( -dY/Y )dY= -ln Y overlimits=-ln(abs(cosb))+ln(abs(cosa))
check my thinking. abs= absolutevalue
so you're saying:
-ln |cosb| + ln |cosa|
I'm still a little confused. Can you help clarify?
int (sinx/cosx)dx
int( -dY/cosx )dY
-ln cosx overlimits
= -ln |cosb| + ln |cosa|
I doubt if I can make it plainer than that, every step is shown. The absolute values is because the log of a negative number is not defined, if that is your question.
Ok. I got it now. Thank you!
To calculate the area under the graph of y = tan(x) between x = a and x = b, we need to use the concept of definite integrals.
The first step is to rewrite tangents as a fraction of sine and cosine. We know that tan(x) = sin(x)/cos(x).
Now, the area under the graph is given by the definite integral of y = tan(x) over the interval [a, b]. The definite integral can be denoted as:
∫[a,b] tan(x) dx
To evaluate this integral, we can convert tan(x) to sin(x)/cos(x):
∫[a, b] sin(x)/cos(x) dx
Next, we can break down the fraction into two separate integrals:
∫[a, b] sin(x) dx / ∫[a, b] cos(x) dx
To calculate each integral, we need to find the antiderivative of sin(x) and cos(x). The antiderivative of sin(x) is -cos(x), and the antiderivative of cos(x) is sin(x).
So now we have:
[-cos(x)] [a, b] / [sin(x)] [a, b]
Evaluating the definite integral, we substitute the upper limit (b) into the antiderivative and subtract the result from the lower limit (a):
[-cos(b) + cos(a)] / [sin(b) - sin(a)]
Now we have the expression for the area under the graph of y = tan(x) between x = a and x = b, which is:
[-cos(b) + cos(a)] / [sin(b) - sin(a)]