The diagram shows a sketch of the graph of the curve $\displaystyle y=\frac{1}{4}x-x^3$ together with the tangent to the curve at the point $A(k, 0)$.
Find the area of the region bounded by the curve and the tangent, giving your answer as a fraction in its lowest terms.
So far,
Equation of tangent: $y = -x/2 -1/4$ Equation of Curve = $\frac{1}{4}x -x^3$
Coordinates of point $A=(-0.5,0)$ Coordinates of point $B=(1,-0.75)$
With that, I found the area of curve using this
Upper limit $= 1$
Lower limit $= -1/2$
$$\int{\frac{1}{4}x-x^3}$$ $$=[x^2/8 - x^4/4]$$ $$=-1/8 - 1/64 = 9/64$$
The solution to this question is $\frac{27}{64}$. How to solve this question?
You are correct in that $A=(-0.5,0)$ and $B=(1,0.75)$. Given the equation of the curve is $\frac{1}{4}x-x^3$, then the equation of the tangent is $-\frac{x}{2}-\frac{1}{4}$. The integral you should evaluate should therefore be:
$$\int_{-0.5}^{1}{(\frac{1}{4}x-x^3)-(-\frac{x}{2}-\frac{1}{4})}$$ $$\int_{-0.5}^{1}{-x^3+\frac{3x}{4}}+\frac{1}{4}$$
You subtract the equation of the tangent from the equation of the curve since the tangent creates a sort of bound to the curve. Your mistake was that you only evaluated the integral of the curve, without considering what the difference evaluating between the tangent and curve would be when the bounds are different. It seems that you are completely able to evaluate these sort of integrals. Can you continue from here to arrive at $\frac{27}{64}$?