What is the area of the region bounded by y=x^2, the tangent to this parabola at (1, 1) and the x-axis?

Question

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?

Answer 1

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.