1. **Problem statement:** Given the derivative of a curve $$\frac{dy}{dx} = a e^{1-x} - 3x^2$$ where $$a$$ is a constant, and the point $$(1,4)$$ on the curve has a gradient of 2. (i) Find the value of $$a$$. (ii) Find the equation of the curve. 2. **Formula and rules:** - The gradient at a point is the value of $$\frac{dy}{dx}$$ at that point. - To find $$a$$, substitute $$x=1$$ and $$\frac{dy}{dx}=2$$ into the derivative. - To find the equation of the curve, integrate $$\frac{dy}{dx}$$ with respect to $$x$$ and use the point $$(1,4)$$ to find the constant of integration. 3. **Step (i) Find $$a$$:** $$\frac{dy}{dx} = a e^{1-x} - 3x^2$$ At $$x=1$$, gradient $$=2$$: $$2 = a e^{1-1} - 3(1)^2 = a e^0 - 3 = a - 3$$ Solve for $$a$$: $$2 = a - 3$$ $$a = 2 + 3 = 5$$ 4. **Step (ii) Find the equation of the curve:** Substitute $$a=5$$: $$\frac{dy}{dx} = 5 e^{1-x} - 3x^2$$ Integrate both sides with respect to $$x$$: $$y = \int (5 e^{1-x} - 3x^2) dx$$ Split the integral: $$y = 5 \int e^{1-x} dx - 3 \int x^2 dx$$ 5. **Integrate each term:** For $$\int e^{1-x} dx$$, let $$u = 1 - x$$, so $$du = -dx$$, thus: $$\int e^{1-x} dx = \int e^u (-du) = - \int e^u du = - e^u + C = - e^{1-x} + C$$ For $$\int x^2 dx$$: $$\int x^2 dx = \frac{x^3}{3} + C$$ 6. **Combine results:** $$y = 5(- e^{1-x}) - 3 \cdot \frac{x^3}{3} + C = -5 e^{1-x} - x^3 + C$$ 7. **Use point $$(1,4)$$ to find $$C$$:** Substitute $$x=1, y=4$$: $$4 = -5 e^{1-1} - 1^3 + C = -5 e^0 - 1 + C = -5 - 1 + C = -6 + C$$ Solve for $$C$$: $$C = 4 + 6 = 10$$ 8. **Final equation of the curve:** $$y = -5 e^{1-x} - x^3 + 10$$ --- 9. **(b)(i) Find $$\int (7x + 8)^{1/3} dx$$:** Let $$u = 7x + 8$$, then $$du = 7 dx$$ or $$dx = \frac{du}{7}$$ Rewrite the integral: $$\int (7x + 8)^{1/3} dx = \int u^{1/3} \frac{du}{7} = \frac{1}{7} \int u^{1/3} du$$ Integrate: $$\int u^{1/3} du = \frac{u^{4/3}}{4/3} = \frac{3}{4} u^{4/3} + C$$ So: $$\int (7x + 8)^{1/3} dx = \frac{1}{7} \cdot \frac{3}{4} u^{4/3} + C = \frac{3}{28} (7x + 8)^{4/3} + C$$ 10. **(b)(ii) Evaluate $$\int_0^8 (7x + 8)^{1/3} dx$$:** Use the antiderivative: $$F(x) = \frac{3}{28} (7x + 8)^{4/3}$$ Evaluate at bounds: $$F(8) - F(0) = \frac{3}{28} (7 \cdot 8 + 8)^{4/3} - \frac{3}{28} (7 \cdot 0 + 8)^{4/3} = \frac{3}{28} (64)^{4/3} - \frac{3}{28} (8)^{4/3}$$ Calculate powers: $$64^{4/3} = (64^{1/3})^4 = 4^4 = 256$$ $$8^{4/3} = (8^{1/3})^4 = 2^4 = 16$$ So: $$\frac{3}{28} (256 - 16) = \frac{3}{28} \times 240 = \frac{720}{28} = \frac{180}{7}$$ **Final answer:** $$\frac{180}{7}$$