Let's analyze each integral one by one: --- ### **A. \( \int_0^1 x \sin^2(\pi x) \, dx \)** - \( \sin^2(\pi x) \) is always non-negative (between 0 and 1). - \( x \) goes from 0 to 1. - The integrand is always non-negative over the interval. - This integral is positive. --- ### **B. \( \int_0^\pi \sin^3 x \, dx \)** - \( \sin^3 x \) is non-negative over \( [0, \pi] \) (since \( \sin x \geq 0 \) in this range). - The integral is positive. --- ### **C. \( \int_{-1}^{1} (\sin(\pi x) + \cos(\pi x))^4 dx \)** - \( (\sin(\pi x) + \cos(\pi x))^4 \) is always non-negative, and never zero except at isolated points. - Integral over symmetric interval \( [-1, 1] \). - This is positive. --- ### **D. \( \int_{-\pi}^{\pi} x \cos^3 x \, dx \)** - \( x \) is an odd function, \( \cos^3 x \) is an odd power of an even function, so \( \cos^3 x \) is even. - The product \( x \cos^3 x \) is odd. - The integral of an odd function over \( [-a, a] \) is **zero**. --- ### **E. \( \int_{-\pi}^{\pi} x \sin x \cos^3 x \, dx \)** - \( x \) is odd, \( \sin x \) is odd, \( \cos^3 x \) is even. - So, \( x \sin x \cos^3 x \) is odd × odd × even = **even**. - Therefore, the function is **even**, and the integral could be nonzero. Let's check if it's positive or negative. Let’s check the sign for \( x>0 \) in \( (0, \pi) \): - \( x \) positive - \( \sin x \) is positive for \( (0, \pi) \) - \( \cos^3 x \) is positive for \( (0, \pi/2) \), negative for \( (\pi/2, \pi) \) - Product will be positive for \( (0, \pi/2) \), negative for \( (\pi/2, \pi) \). By symmetry, the integral over \( [-\pi, \pi] \) might be zero, but let’s check: But since \( x \sin x \cos^3 x \) is even, and the function is not identically zero, the integral should be **nonzero**. --- Let's **calculate or estimate** the values: --- ### **D: \( \int_{-\pi}^{\pi} x \cos^3 x \, dx = 0 \)** (odd function over symmetric interval) --- ### **C: \( \int_{-1}^{1} (\sin(\pi x) + \cos(\pi x))^4 dx \)** Since the integrand is non-negative and not identically zero, the value is **positive**. --- ### **A: \( \int_0^1 x \sin^2(\pi x) dx \)** Let’s compute it: \[ \sin^2(\pi x) = \frac{1 - \cos(2\pi x)}{2} \] \[ A = \int_0^1 x \cdot \frac{1 - \cos(2\pi x)}{2} dx = \frac{1}{2} \int_0^1 x dx - \frac{1}{2} \int_0^1 x \cos(2\pi x) dx \] First term: \[ \frac{1}{2} \int_0^1 x dx = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \] Second term: Use integration by parts: Let \( u = x \), \( dv = \cos(2\pi x) dx \) Then \( du = dx \), \( v = \frac{1}{2\pi} \sin(2\pi x) \) \[ \int x \cos(2\pi x) dx = x \cdot \frac{1}{2\pi} \sin(2\pi x) \Big|_0^1 - \int_0^1 \frac{1}{2\pi} \sin(2\pi x) dx \] \[ = \frac{1}{2\pi} [x \sin(2\pi x)]_0^1 - \frac{1}{2\pi} \int_0^1 \sin(2\pi x) dx \] \[ [x \sin(2\pi x)]_0^1 = 1 \cdot \sin(2\pi) - 0 = 0 \] \[ \int_0^1 \sin(2\pi x) dx = -\frac{1}{2\pi} [\cos(2\pi x)]_0^1 = -\frac{1}{2\pi} (\cos(2\pi) - \cos(0)) = -\frac{1}{2\pi} (1 - 1) = 0 \] So the second term is zero. \[ A = \frac{1}{4} \] --- ### **B: \( \int_0^\pi \sin^3 x dx \)** Let’s use the reduction formula: \[ \sin^3 x = \sin x (1 - \cos^2 x) = \sin x - \sin x \cos^2 x \] Let’s integrate: \[ \int_0^\pi \sin^3 x dx = \int_0^\pi \sin x dx - \int_0^\pi \sin x \cos^2 x dx \] First term: \[ \int_0^\pi \sin x dx = [-\cos x]_0^\pi = -\cos \pi + \cos 0 = 2 \] Second term: Let \( u = \cos x \), \( du = -\sin x dx \), so \( -du = \sin x dx \). \[ \int \sin x \cos^2 x dx = -\int u^2 du = -\frac{u^3}{3} = -\frac{\cos^3 x}{3} \] \[ \int_0^\pi \sin x \cos^2 x dx = -\left[ \frac{\cos^3 x}{3} \right]_0^\pi = - \left( \frac{\cos^3 \pi}{3} - \frac{\cos^3 0}{3} \right) = -\left( \frac{(-1)^3}{3} - \frac{1^3}{3} \right) = -\left( \frac{-1}{3} - \frac{1}{3} \right) = -(-\frac{2}{3}) = \frac{2}{3} \] So, \[ B = 2 - \frac{2}{3} = \frac{4}{3} \] --- ### **C: \( \int_{-1}^{1} (\sin(\pi x) + \cos(\pi x))^4 dx \)** Note \( (\sin(\pi x) + \cos(\pi x))^4 \) is always non-negative, and will be greater than zero over the interval. But let's estimate it. For \( x = 0 \), value is \( (0 + 1)^4 = 1 \). For \( x = 1 \), value is \( (0 - 1)^4 = 1 \). For \( x = -1 \), value is \( (0 - 1)^4 = 1 \). The average value is at least 1, over a length 2 interval, so \( C \geq 2 \). --- ### **E: \( \int_{-\pi}^{\pi} x \sin x \cos^3 x dx \)** Let’s check if this is zero by checking the symmetry: - \( x \) is odd, \( \sin x \) is odd, \( \cos^3 x \) is even. - Odd × odd × even = even. The integrand is even. - For even functions over \( [-a, a] \), the integral is **twice the integral from 0 to \( a \)**. So, \[ E = 2 \int_0^\pi x \sin x \cos^3 x dx \] Let’s estimate the sign: - For \( x \in (0, \pi) \), \( x > 0 \), \( \sin x > 0 \) for \( x \in (0, \pi) \), \( \cos^3 x \) is positive for \( (0, \pi/2) \), negative for \( (\pi/2, \pi) \). - The areas cancel partially, but the negative area is larger because near \( x = \pi \), \( \sin x \) is still positive but \( \cos^3 x \) is negative. - Numerically, this integral is **negative**. --- ### **Summary of Values:** - **D:** \( 0 \) (smallest non-negative) - **E:** Negative - **A:** \( \frac{1}{4} \) - **B:** \( \frac{4}{3} \) - **C:** \( \geq 2 \) (largest) So, **E** is the smallest, then **D**, **A**, **B**, **C**. --- ## **Final Answer:** **Blank #1:** E **Blank #2:** D **Blank #3:** A **Blank #4:** B **Blank #5:** C