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 problem, we need to find the value of

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

where \( f(x) = 4x^3 + 2x - 6 \).

### Step 1: Calculate \(\int_{0}^{2} f(x) \, dx\)

We first 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 the integral term by term:

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

Calculating each integral:

1. \(\int 4x^3 \, dx = x^4\), so:
\[ \left[ x^4 \right]_{0}^{2} = 2^4 - 0^4 = 16 \]

2. \(\int 2x \, dx = x^2\), so:
\[ \left[ x^2 \right]_{0}^{2} = 2^2 - 0^2 = 4 \]

3. \(\int 6 \, dx = 6x\), so:
\[ \left[ 6x \right]_{0}^{2} = 6(2) - 6(0) = 12 \]

Putting it all together:

\[ \int_{0}^{2} f(x) \, dx = 16 + 4 - 12 = 8 \]

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

Using the property of integrals and inverse functions, we have:

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

In our case, \( a = 0 \) and \( b = 30 \). We need to find \( f^{-1}(0) \) and \( f^{-1}(30) \).

1. **Finding \( f^{-1}(0) \)**:
Set \( f(x) = 0 \):
\[ 4x^3 + 2x - 6 = 0 \]
Testing \( x = 1 \):
\[ f(1) = 4(1)^3 + 2(1) - 6 = 4 + 2 - 6 = 0 \]
Thus, \( f^{-1}(0) = 1 \).

2. **Finding \( f^{-1}(30) \)**:
Set \( f(x) = 30 \):
\[ 4x^3 + 2x - 6 = 30 \implies 4x^3 + 2x - 36 = 0 \]
Testing \( x = 2 \):
\[ f(2) = 4(2)^3 + 2(2) - 6 = 32 + 4 - 6 = 30 \]
Thus, \( f^{-1}(30) = 2 \).

Now we can apply the property:

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

Calculating the right-hand side:

\[ 30 \cdot 2 = 60 \]

Thus,

\[ \int_{0}^{30} f^{-1}(y) \, dy + \int_{1}^{2} f(x) \, dx = 60 \]

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

We already calculated \( \int_{0}^{2} f(x) \, dx = 8 \), so:

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

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

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

Calculating each integral:

1. \(\int 4x^3 \, dx = x^4\), so:
\[ \left[ x^4 \right]_{0}^{1} = 1 - 0 = 1 \]

2. \(\int 2x \, dx = x^2\), so:
\[ \left[ x^2 \right]_{0}^{1} = 1 - 0 = 1 \]

3. \(\int 6 \, dx = 6x\), so:
\[ \left[ 6x \right]_{0}^{1} = 6(1) - 6(0) = 6 \]

Putting it all together:

\[ \int_{0}^{1} f(x) \, dx = 1 + 1 - 6 = -4 \]

Thus,

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

### Step 4: Substitute back into the equation

Now substituting back into our earlier equation:

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

So,

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

### Final Calculation

Finally, we add both integrals:

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

Thus, the final answer is:

\[ \boxed{56} \]