To solve the problem, we need to evaluate the expression:
\[
\int_{0}^{2} f(x) \, dx + \int_{0}^{30} f^{-1}(y) \, dy
\]
where \( f(x) = 4x^3 + 2x - 6 \).
### Step 1: Calculate \( \int_{0}^{2} f(x) \, dx \)
First, we need to compute the integral of \( f(x) \) from 0 to 2.
\[
\int_{0}^{2} (4x^3 + 2x - 6) \, dx
\]
We can integrate each term separately:
\[
\int 4x^3 \, dx = \frac{4}{4} x^4 = x^4
\]
\[
\int 2x \, dx = x^2
\]
\[
\int -6 \, dx = -6x
\]
Now, we can combine these results:
\[
\int (4x^3 + 2x - 6) \, dx = x^4 + x^2 - 6x
\]
Now we evaluate this from 0 to 2:
\[
\left[ x^4 + x^2 - 6x \right]_{0}^{2} = \left[ 2^4 + 2^2 - 6 \cdot 2 \right] - \left[ 0^4 + 0^2 - 6 \cdot 0 \right]
\]
Calculating the upper limit:
\[
= 16 + 4 - 12 = 8
\]
The lower limit is 0, so:
\[
\int_{0}^{2} f(x) \, dx = 8
\]
### Step 2: Find \( f^{-1}(y) \) and calculate \( \int_{0}^{30} f^{-1}(y) \, dy \)
To find \( f^{-1}(y) \), we need to find the values of \( x \) such that \( f(x) = y \).
First, we need to find \( f(1) \) and \( f(2) \):
\[
f(1) = 4(1)^3 + 2(1) - 6 = 4 + 2 - 6 = 0 \quad \Rightarrow \quad f^{-1}(0) = 1
\]
\[
f(2) = 4(2)^3 + 2(2) - 6 = 32 + 4 - 6 = 30 \quad \Rightarrow \quad f^{-1}(30) = 2
\]
Now we can change the variable in the second integral:
Using the substitution \( y = f(t) \), we have \( dy = f'(t) \, dt \). The limits change from \( y = 0 \) to \( y = 30 \), which corresponds to \( t = 1 \) to \( t = 2 \).
Thus, we can rewrite the integral:
\[
\int_{0}^{30} f^{-1}(y) \, dy = \int_{1}^{2} t \cdot f'(t) \, dt
\]
Now we calculate \( f'(t) \):
\[
f'(x) = 12x^2 + 2
\]
So the integral becomes:
\[
\int_{1}^{2} t (12t^2 + 2) \, dt = \int_{1}^{2} (12t^3 + 2t) \, dt
\]
Integrating term by term:
\[
\int 12t^3 \, dt = 3t^4
\]
\[
\int 2t \, dt = t^2
\]
Thus,
\[
\int (12t^3 + 2t) \, dt = 3t^4 + t^2
\]
Evaluating from 1 to 2:
\[
\left[ 3t^4 + t^2 \right]_{1}^{2} = \left[ 3(2^4) + (2^2) \right] - \left[ 3(1^4) + (1^2) \right]
\]
\[
= \left[ 3(16) + 4 \right] - \left[ 3(1) + 1 \right] = (48 + 4) - (3 + 1) = 52 - 4 = 48
\]
### Step 3: Combine the results
Now we can combine the results of both integrals:
\[
\int_{0}^{2} f(x) \, dx + \int_{0}^{30} f^{-1}(y) \, dy = 8 + 48 = 56
\]
Thus, the final answer is:
\[
\boxed{56}
\]
