The curve C is defined for t ≥ 0 by the parametric equations $x = t^2 + t$ and $y = 4t^2 - t^3$ C is shown in the diagram below - AQA - A-Level Maths Pure - Question 14 - 2021 - Paper 1

To find b, note that the x-coordinate of the point where the curve intersects the x-axis is at $t=4$. The corresponding value for x is:

$x = 4^2 + 4 = 20.$

Thus, the value of b is 20.

Using the substitution $y = 4t^2 - t^3$, we express the area A as:

$A = \int_0^{b} y \, dx = \int_0^{4} (4t^2 - t^3) (2t + 1) \, dt.$

This results in:

$A = \int_0^4 (4t^2(2t + 1) - t^3(2t + 1)) \, dt$

This simplifies to:

$A = \int_0^4 (8t^3 + 4t^2 - 2t^4 - t^4) \, dt$

or:

$A = \int_0^4 (8t^3 + 4t^2 - 3t^4) \, dt.$

On integrating gives:

$A = \int_0^{4} (4t^2 + 7t^3 - 2t^4) \, dt$

To compute the area A, evaluate:

$A = \int_0^{4} (4t^2 + 7t^3 - 2t^4) \ dt.$

Calculating each term separately:

1. For $4t^2$: $\int 4t^2 dt = \frac{4}{3}t^3$
2. For $7t^3$: $\int 7t^3 dt = \frac{7}{4}t^4$
3. For $-2t^4$: $\int -2t^4 dt = -\frac{2}{5}t^5$

Finally, substituting limits:

$A = \left[\frac{4}{3}(4^3) + \frac{7}{4}(4^4) - \frac{2}{5}(4^5)\right] - \left[0\right].$

Calculating each term:

1. $\frac{4}{3}(64) = \frac{256}{3}$
2. $\frac{7}{4}(256) = 448$
3. $-\frac{2}{5}(1024) = -\frac{2048}{5}$

Now combine:

Converting to a common denominator will yield:

$\frac{2560}{15} + \frac{6720}{15} - \frac{6144}{15} = \frac{4036}{15}.$