Question

1. $\frac{dy}{dx}=\sin x - e^{-x}+8x^{3}$

Explanation:

Step1: Integrate each term separately

$$y=\int(\sin x - e^{-x}+8x^{3})dx=\int\sin xdx-\int e^{-x}dx + 8\int x^{3}dx$$

Step2: Apply integration formulas

• For $\int\sin xdx$, using the formula $\int\sin xdx=-\cos x + C_1$.
• For $\int e^{-x}dx$, let $u = -x$, then $du=-dx$. So $\int e^{-x}dx=-\int e^{u}du=-e^{u}+C_2=-e^{-x}+C_2$.
• For $8\int x^{3}dx$, using the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n

eq - 1)$, we have $8\int x^{3}dx=8\times\frac{x^{4}}{4}+C_3 = 2x^{4}+C_3$.

Step3: Combine the results

$$y=-\cos x+e^{-x}+2x^{4}+C$$ (where $C = C_1 + C_2+C_3$)

Answer:

$y =-\cos x+e^{-x}+2x^{4}+C$