Use symmetry to evaluate the following integral. \( \int_{-1}^{1}\left(6 x^{4}-7\right) d x \) \( \int_{-1}^{1}\left(6 x^{4}-7\right) d x=\square \) (Type an integer or a simplified fraction.)

We begin by recognizing that the function \(6x^4\) and the constant \(-7\) are both even functions (i.e., \(f(x) = f(-x)\)). For an even function \(f(x)\), the integral over the interval \([-a, a]\) can be written as \[ \int_{-a}^{a} f(x)\, dx = 2\int_{0}^{a} f(x)\, dx. \] Thus, we have \[ \int_{-1}^{1} \left(6x^4 - 7\right) dx = 2\int_{0}^{1} \left(6x^4 - 7\right) dx. \] Next, we split the integral: \[ 2\int_{0}^{1} \left(6x^4 - 7\right) dx = 2\left(6\int_{0}^{1} x^4\, dx - 7\int_{0}^{1} dx \right). \] Now we evaluate each definite integral separately. 1. Evaluate \(\int_{0}^{1} x^4\, dx\): \[ \int_{0}^{1} x^4\, dx = \left[\frac{x^5}{5}\right]_{0}^{1} = \frac{1^5}{5} - \frac{0^5}{5} = \frac{1}{5}. \] 2. Evaluate \(\int_{0}^{1} dx\): \[ \int_{0}^{1} dx = \left[x\right]_{0}^{1} = 1 - 0 = 1. \] Substitute these back into the expression: \[ 2\left(6\cdot\frac{1}{5} - 7\cdot 1\right) = 2\left(\frac{6}{5} - 7\right). \] Simplify inside the parentheses: \[ \frac{6}{5} - 7 = \frac{6 - 35}{5} = \frac{-29}{5}. \] Finally, multiply by 2: \[ 2\left(\frac{-29}{5}\right) = \frac{-58}{5}. \] Thus, the value of the integral is \[ \boxed{\frac{-58}{5}}. \]