Find the area of the region bounded by C: y = tanx , tangent drawn at C at x = pie/4 and the x- axis.
To find the area of the region bounded by the curve C: y = tan(x), the tangent line drawn at C at x = π/4, and the x-axis, we can use the method of integration.
Step 1: Find the x-coordinate of the intersection point between the curve C and the tangent line.
Substituting x = π/4 into the equation of C:
y = tan(π/4) = 1
So the point of intersection is (π/4, 1).
Step 2: Set up the definite integral that represents the area bounded by C, the tangent line, and the x-axis.
The area bounded by the curve C, the tangent line, and the x-axis can be represented by the integral:
∫[a,b] (f(x) - g(x)) dx,
where f(x) is the function for the curve C, g(x) is the function for the x-axis (g(x) = 0), and [a, b] is the interval of integration.
In this case, the interval of integration is [0, π/4], and f(x) = tan(x) and g(x) = 0.
So the integral becomes:
∫[0, π/4] (tan(x) - 0) dx.
Step 3: Evaluate the integral.
∫[0, π/4] tan(x) dx = -ln|cos(x)| from 0 to π/4.
Evaluating this integral, we have:
[-ln|cos(π/4)|] - [-ln|cos(0)|].
Since cos(0) = 1 and cos(π/4) = √2/2, the integral becomes:
[-ln(√2/2)] - [-ln(1)].
Simplifying further:
[-ln(√2/2)] - [0] = -ln(√2/2).
So, the area of the region bounded by C, the tangent line, and the x-axis is -ln(√2/2), or approximately 0.2027 square units.
To find the area of the region bounded by the curve C: y = tan(x), the tangent line at C when x = π/4, and the x-axis, you can follow these steps:
1. Identify the limits of integration: To determine the boundaries of the region, we need to find the x-values where the curve intersects the tangent line and the x-axis.
- The curve C: y = tan(x) intersects the x-axis when y = 0, so we need to solve the equation tan(x) = 0. This occurs at x = 0, π, 2π, etc.
- The tangent line at C intersects the x-axis when y = 0. To find the x-value, we solve tan(x) = tan(π/4). The solution is x = π/4 + kπ, where k is any integer.
Therefore, the limits of integration for x are 0 and π/4.
2. Setup the integral: Since we are finding the area, we need to integrate the absolute value of the curve's function.
The area can be calculated as: A = ∫[0, π/4] |tan(x)| dx
3. Evaluate the integral: To evaluate the integral, integrate the absolute value of tan(x) within the given limits of integration.
A = ∫[0, π/4] tan(x) dx
You can use various methods, such as integration by parts or a substitution, to calculate this integral. After evaluating the integral, you will obtain the value of the area.
Please note that the given integral might not have a closed-form solution, so you may need to use numerical methods or approximation techniques to find the area accurately.
We will need the equation of the tangent.
y = tanx
y' = sec^2 x
when x = π/4 , tan π/4 = 1
so at the point (π/4,1)
the slope is sec^2 (π/4) = 2
so equation is y = 2x + b
at (π/4,1)
1 = 2(π/4) + b
b = 1 - π/2 = (2-π)/2
tangent equation is y = 2x + (2-π)/2
the x-intercept, let y = 0
2x = (π-2)/2 , x = (π-2)/4
I hope you made a sketch
I am going to take the area below y = tanx from 0 to π/4, then subtract the little right-angled triangle.
first part of area
= ∫tanx dx from 0 to π/4
= -ln(cosx) | from 0 to π/4
= (-ln(1/√2) - (-ln(1) )
= - ( ln1- ln√2) + 0
= ln √2
base of little triangle = π/4 - (π-2)/4
= (π - π + 2)/4 = 1/2
height = 1
area = (1/2(1/2)(1) = 1/4
area you want =ln(√2) - 1/4 or
= (4ln√2 - 1)/4
= 2ln2 - 1)/4 or
= (ln4 - 1)/4