Integration — practice questions

To find the equation of the curve, we must integrate the gradient function. $y = \int (3x^2 - 8x + 1) \, dx$

Apply the power rule for integration to each term: $y = \frac{3x^3}{3} - \frac{8x^2}{2} + x + C$ $y = x^3 - 4x^2 + x + C$

This is the general equation for the family of curves with the given gradient. To find the specific curve, we use the point $(2, -5)$. Substitute $x=2$ and $y=-5$: $-5 = (2)^3 - 4(2)^2 + (2) + C$ $-5 = 8 - 4(4) + 2 + C$ $-5 = 8 - 16 + 2 + C$ $-5 = -6 + C$ $C = 1$

Therefore, the specific equation of the curve is: $y = x^3 - 4x^2 + x + 1$

First, we need to find the limits of integration. The region is bounded by the x-axis, so we find the x-intercepts by setting $y=0$. $4 - x^2 = 0$ $x^2 = 4$ $x = -2$ and $x = 2$ So, our limits are $a=-2$ and $b=2$.

The area is given by the definite integral: Area $= \int_{-2}^{2} (4 - x^2) \, dx$

First, integrate the function: $[4x - \frac{x^3}{3}]_{-2}^{2}$

Now, evaluate at the upper limit ($x=2$) and subtract the evaluation at the lower limit ($x=-2$). This step is crucial for method marks. $= (4(2) - \frac{(2)^3}{3}) - (4(-2) - \frac{(-2)^3}{3})$ $= (8 - \frac{8}{3}) - (-8 - \frac{-8}{3})$ $= (8 - \frac{8}{3}) - (-8 + \frac{8}{3})$ $= (\frac{24}{3} - \frac{8}{3}) - (-\frac{24}{3} + \frac{8}{3})$ $= (\frac{16}{3}) - (-\frac{16}{3})$ $= \frac{16}{3} + \frac{16}{3} = \frac{32}{3}$

The exact area is $\frac{32}{3}$ square units.