Step-by-step Solution:

Statement I: \[ \int_{-a}^{a} f(x) \, dx = \int_{0}^{a} [f(x) + f(-x)] \, dx \] Let's verify this statement by considering the properties of integrals: 1. Splitting the integral: \[ \int_{-a}^{a} f(x) \, dx = \int_{-a}^{0} f(x) \, dx + \int_{0}^{a} f(x) \, dx \] 2. Changing the variable in the first integral: Let \( x = -u \), then \( dx = -du \), and when \( x = -a \), \( u = a \); when \( x = 0 \), \( u = 0 \). So, \[ \int_{-a}^{0} f(x) \, dx = \int_{a}^{0} f(-u) (-du) = \int_{0}^{a} f(-u) \, du \] 3. Combining the integrals: \[ \int_{-a}^{a} f(x) \, dx = \int_{0}^{a} f(-u) \, du + \int_{0}^{a} f(x) \, dx = \int_{0}^{a} [f(x) + f(-x)] \, dx \] Statement I is correct. Statement II: \[ \int_{0}^{1} \sqrt{(1+x)(1+x^3)} \, dx \leq \frac{15}{8} \] To verify this, let's estimate the integral: 1. Simplify the integrand: \[ \sqrt{(1+x)(1+x^3)} = \sqrt{1 + x + x^3 + x^4} \] 2. Estimate the maximum value of the integrand on the interval [0, 1]: At \( x = 1 \): \[ \sqrt{(1+1)(1+1)} = \sqrt{4} = 2 \] At \( x = 0 \): \[ \sqrt{(1+0)(1+0)} = \sqrt{1} = 1 \] The function \( \sqrt{(1+x)(1+x^3)} \) increases on [0, 1], so its maximum value is 2. 3. Estimate the integral: \[ \int_{0}^{1} \sqrt{(1+x)(1+x^3)} \, dx \leq \int_{0}^{1} 2 \, dx = 2 \] However, \( \frac{15}{8} = 1.875 \), which is less than 2. Therefore, the statement that the integral is less than or equal to \( \frac{15}{8} \) is not necessarily true based on this estimation. Statement II is incorrect. Conclusion: - Statement I is correct. - Statement II is incorrect. Therefore, the correct answer is: C. Statement I is correct but Statement II is incorrect