Let `y=f(x)=4x^(3)+2x-6`, then the value of `int_(0)^(2)f(x)dx+int_(0)^(30)f^(-1)(y)dy` is equal to _________.

Explanation

To solve the given 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: Evaluate \( \int_{0}^{2} f(x) \, dx \)

First, we will compute the integral of \( f(x) \) from 0 to 2.

\[ \int_{0}^{2} f(x) \, dx = \int_{0}^{2} (4x^3 + 2x - 6) \, dx \]

Calculating this integral:

\[ = \left[ \frac{4}{4}x^4 + \frac{2}{2}x^2 - 6x \right]_{0}^{2} \]

\[ = \left[ x^4 + x^2 - 6x \right]_{0}^{2} \]

Now, substituting the limits:

\[ = \left[ 2^4 + 2^2 - 6 \cdot 2 \right] - \left[ 0^4 + 0^2 - 6 \cdot 0 \right] \]

\[ = \left[ 16 + 4 - 12 \right] - 0 \]

\[ = 16 + 4 - 12 = 8 \]

### Step 2: Evaluate \( \int_{0}^{30} f^{-1}(y) \, dy \)

Using the property of inverse functions, we know that:

\[ \int_{a}^{b} f^{-1}(y) \, dy = b f^{-1}(b) - a f^{-1}(a) - \int_{f^{-1}(a)}^{f^{-1}(b)} f(x) \, dx \]

We need to find \( f^{-1}(0) \) and \( f^{-1}(30) \).

#### Finding \( f^{-1}(0) \)

Set \( f(x) = 0 \):

\[ 4x^3 + 2x - 6 = 0 \]

By testing \( x = 1 \):

\[ 4(1)^3 + 2(1) - 6 = 4 + 2 - 6 = 0 \]

Thus, \( f^{-1}(0) = 1 \).

#### Finding \( f^{-1}(30) \)

Set \( f(x) = 30 \):

\[ 4x^3 + 2x - 6 = 30 \]

This simplifies to:

\[ 4x^3 + 2x - 36 = 0 \]

By testing \( x = 2 \):

\[ 4(2)^3 + 2(2) - 36 = 32 + 4 - 36 = 0 \]

Thus, \( f^{-1}(30) = 2 \).

Now substituting into the integral:

\[ \int_{0}^{30} f^{-1}(y) \, dy = 30 \cdot f^{-1}(30) - 0 \cdot f^{-1}(0) - \int_{1}^{2} f(x) \, dx \]

This simplifies to:

\[ = 30 \cdot 2 - 0 - \int_{1}^{2} f(x) \, dx \]

### Step 3: Evaluate \( \int_{1}^{2} f(x) \, dx \)

We already computed \( \int_{0}^{2} f(x) \, dx \) and can find \( \int_{1}^{2} f(x) \, dx \):

\[ \int_{1}^{2} f(x) \, dx = \int_{0}^{2} f(x) \, dx - \int_{0}^{1} f(x) \, dx \]

Calculating \( \int_{0}^{1} f(x) \, dx \):

\[ \int_{0}^{1} (4x^3 + 2x - 6) \, dx = \left[ x^4 + x^2 - 6x \right]_{0}^{1} \]

\[ = \left[ 1 + 1 - 6 \right] - 0 = -4 \]

Thus:

\[ \int_{1}^{2} f(x) \, dx = 8 - (-4) = 12 \]

### Step 4: Substitute back into the integral for \( f^{-1}(y) \)

Now we can substitute back:

\[ \int_{0}^{30} f^{-1}(y) \, dy = 30 \cdot 2 - 12 = 60 - 12 = 48 \]

### Step 5: Combine results

Finally, we combine both integrals:

\[ \int_{0}^{2} f(x) \, dx + \int_{0}^{30} f^{-1}(y) \, dy = 8 + 48 = 56 \]

Thus, the final answer is:

\[ \boxed{56} \]