To find $g(x)$, given $g'(x) = (60x^2 - 80x) \cdot (74x^3 - 8x^2)$ and $g(0) = 1$, we need to integrate $g'(x)$ with respect to $x$ and then apply the initial condition $g(0) = 1$.

Step 1: Simplify $g'(x)$

Rewrite $g'(x)$: $g'(x) = (60x^2 - 80x)(74x^3 - 8x^2)$

Expand $g'(x)$ by multiplying the terms: $g'(x) = (60x^2 \cdot 74x^3) - (60x^2 \cdot 8x^2) - (80x \cdot 74x^3) + (80x \cdot 8x^2)$

This gives: $g'(x) = 4440x^5 - 480x^4 - 5920x^4 + 640x^3$ Combine like terms: $g'(x) = 4440x^5 - 6400x^4 + 640x^3$

Step 2: Integrate $g'(x)$ to find $g(x)$

Now, we integrate $g'(x)$ with respect to $x$: $g(x) = \int (4440x^5 - 6400x^4 + 640x^3) \, dx$

Integrating each term separately:

1. $\int 4440x^5 \, dx = \frac{4440}{6}x^6 = 740x^6$
2. $\int -6400x^4 \, dx = -\frac{6400}{5}x^5 = -1280x^5$
3. $\int 640x^3 \, dx = \frac{640}{4}x^4 = 160x^4$

Thus, $g(x) = 740x^6 - 1280x^5 + 160x^4 + C$

Step 3: Apply the initial condition $g(0) = 1$

Substitute $x = 0$ and $g(0) = 1$: $g(0) = 740 \cdot 0^6 - 1280 \cdot 0^5 + 160 \cdot 0^4 + C = 1$ This simplifies to $C = 1$.

Final Answer

Therefore, $g(x) = 740x^6 - 1280x^5 + 160x^4 + 1$

Would you like more details on any part of the solution?

Here are five related questions to consider:

1. How would the solution change if $g(0)$ had a different value?
2. How can you check the correctness of $g(x)$ by differentiating it?
3. What techniques can be used to simplify products of polynomials before integrating?
4. How would the solution differ if there were additional terms in $g'(x)$?
5. How does the degree of $g(x)$ relate to the degree of $g'(x)$?

Tip: Always remember to apply initial conditions after integrating to determine the constant of integration accurately.