1. ## Definite integral

The diagram above shows the straight line $y=4x+1$ and the curve $y=(2x-3)(x-2)$ intersecting at point $(k,3)$.Find

a)the value of $k$

b)the area of the shaded region.

2. Originally Posted by mastermin346
The diagram above shows the straight line $y=4x+1$ and the curve $y=(2x-3)(x-2)$ intersecting at point $(k,3)$.Find

a)the value of $k$

b)the area of the shaded region.
The graphs intersect where they are equal.

So $4x +1 = (2x - 3)(x - 2)$

$4x + 1 = 2x^2 - 4x - 3x + 6$

$4x + 1 = 2x^2 - 7x + 6$

$0 = 2x^2 - 11x + 5$

$0 = 2x^2 - x - 10x + 5$

$0 = x(2x - 1) - 5(2x - 1)$

$0 = (2x - 1)(x - 5)$

$2x - 1 = 0$ or $x - 5 = 0$

$x = \frac{1}{2}$ or $x = 5$.

Clearly the point you are referring to is $(x, y) = \left(\frac{1}{2}, 3\right)$.

So you are wanting to find

$\int_0^{\frac{1}{2}}{2x^2 - 7x + 6\,dx} - \int_0^{\frac{1}{2}}{4x + 1\,dx}$.