What is the area of the region bounded by y=x^2, the tangent to this parabola at (1, 1) and the x-axis?
Since it says that the parabola passes through 1,1 can I assume that the line is y = x or is that completely wrong?
Doing that I got ∫ x - x^2 dx (where the interval from a to b goes from 0 to 1)
so the solved integral would be
x^2/2 - x^3/3] 0 to 1
= [(1)^2/2 - (1)3/3] - 0
= 1/2 - 1/3
= 1/6
What should I have done if it's wrong?
The tangent line is not y = x
Since the derivative of y = x^2 is dy/dx = 2x, at (1,1) dy/dx = 2, so the tangent line has a slope of 2
its equation would be y = 2x + b, with (1,1) on it
So, 1 = 2(1) + b, ------> b = -1
the tangent equation is y = 2x-1, which would result in an x-intercept of (1/2,0).
I would take the area between y = x^2 from 0 to 1 minus the right -angled triangle formed by the x-axis, y=2x-1 and x=1
let me know if you got 1/12.
To find the area of the region bounded by y = x^2, the tangent line at (1, 1), and the x-axis, you need to first find the points where the tangent line intersects the parabola.
Since the tangent line passes through (1, 1), we can write the equation of the tangent line using the point-slope form, y - y1 = m(x - x1), where m is the slope of the tangent line.
To find the slope of the tangent line, we take the derivative of the parabola y = x^2 and evaluate it at x = 1:
dy/dx = 2x
At x = 1, dy/dx = 2(1) = 2.
So the slope of the tangent line is 2.
Using the point-slope form, we can now write the equation of the tangent line:
y - 1 = 2(x - 1)
y - 1 = 2x - 2
y = 2x - 1
To find the points of intersection between the parabola and the tangent line, we set the y-values equal to each other:
x^2 = 2x - 1
This is a quadratic equation that can be rearranged into standard form:
x^2 - 2x + 1 = 0
Now, to find the area of the region, we need to integrate the difference in the two functions y = x^2 and y = 2x - 1, over the interval of x-values where they intersect.
To find the x-values of intersection, we solve the quadratic equation by factoring or using the quadratic formula:
(x - 1)^2 = 0
x - 1 = 0
x = 1
So the parabola and the tangent line intersect at x = 1.
Now we can calculate the area by integrating the difference:
A = ∫[x^2 - (2x - 1)] dx from 0 to 1
Evaluating the integral, we get:
A = [(x^3/3 - x^2/2) - (x^2 - x)] from 0 to 1
A = [(1/3 - 1/2) - (1 - 1)] - [(0/3 - 0/2) - (0 - 0)]
A = [(1/6) - (0)] - [(0) - (0)]
A = 1/6 - 0
A = 1/6
So the area of the region bounded by y = x^2, the tangent line at (1, 1), and the x-axis is 1/6, or approximately 0.1667.