y1= sin (x)
y2= (tanx)
It is graphed in the interval [0, pi/2]
find the integral
To find the integral of a function, we need to evaluate the definite integral over the given interval [0, π/2]. In this case, we have two functions, y1 = sin(x) and y2 = tan(x).
Let's start with y1 = sin(x):
Step 1: Integrate sin(x) with respect to x.
∫ sin(x) dx = -cos(x) + C, where C is the constant of integration.
Step 2: Evaluate the definite integral on the interval [0, π/2].
∫[0, π/2] sin(x) dx = [-cos(x)] [0, π/2]
= -cos(π/2) - (-cos(0))
= -0 - (-1)
= 1.
So, the definite integral of y1 = sin(x) over the interval [0, π/2] is 1.
Now, let's move on to y2 = tan(x):
Step 1: Integrate tan(x) with respect to x.
∫ tan(x) dx = -ln|cos(x)| + C, where C is the constant of integration.
Step 2: Evaluate the definite integral on the interval [0, π/2].
∫[0, π/2] tan(x) dx = [-ln|cos(x)|] [0, π/2].
(Note: The natural logarithm of a negative number is undefined, so we need to consider the absolute value of the cosine function.)
= -ln|cos(π/2)| - (-ln|cos(0)|)
= -ln|0| - (-ln|1|)
= -ln(0) + ln(1)
= -∞ + 0
= -∞.
Therefore, the definite integral of y2 = tan(x) over the interval [0, π/2] is -∞.
To summarize, the integral of y1 = sin(x) over [0, π/2] is 1, while the integral of y2 = tan(x) over the same interval is -∞.