Let's solve the integral step-by-step **with respect to \( s \)**, treating \( t \) as a constant, and then **evaluate with the limits** from \( s = -1 \) to \( s = 1 \): \[ \int_{-1}^{1} \left( 4s^2 t^4 + 8s^2 t^3 + 4s t^4 + 8s t^3 + 4s^2 t^2 + 4s t^2 + t^4 + 2t^3 + t^2 \right) ds \] ### 1. Integrate Each Term with Respect to \( s \): - \( \int 4s^2 t^4 \, ds = 4t^4 \int s^2 \, ds = 4t^4 \cdot \frac{s^3}{3} = \frac{4}{3} t^4 s^3 \) - \( \int 8s^2 t^3 \, ds = 8t^3 \int s^2 \, ds = 8t^3 \cdot \frac{s^3}{3} = \frac{8}{3} t^3 s^3 \) - \( \int 4s t^4 \, ds = 4t^4 \int s \, ds = 4t^4 \cdot \frac{s^2}{2} = 2t^4 s^2 \) - \( \int 8s t^3 \, ds = 8t^3 \int s \, ds = 8t^3 \cdot \frac{s^2}{2} = 4t^3 s^2 \) - \( \int 4s^2 t^2 \, ds = 4t^2 \int s^2 \, ds = 4t^2 \cdot \frac{s^3}{3} = \frac{4}{3} t^2 s^3 \) - \( \int 4s t^2 \, ds = 4t^2 \int s \, ds = 4t^2 \cdot \frac{s^2}{2} = 2t^2 s^2 \) - \( \int t^4 \, ds = t^4 s \) - \( \int 2t^3 \, ds = 2t^3 s \) - \( \int t^2 \, ds = t^2 s \) ### 2. Combine All Together: \[ F(s) = \left( \frac{4}{3} t^4 s^3 + \frac{8}{3} t^3 s^3 + 2t^4 s^2 + 4t^3 s^2 + \frac{4}{3} t^2 s^3 + 2t^2 s^2 + t^4 s + 2t^3 s + t^2 s \right) \] ### 3. Evaluate from \( s = -1 \) to \( s = 1 \): \[ \text{Integral} = F(1) - F(-1) \] Let’s compute \( F(1) \): - \( s^3 = 1 \), \( s^2 = 1 \), \( s = 1 \): \[ F(1) = \left( \frac{4}{3} t^4 + \frac{8}{3} t^3 + 2t^4 + 4t^3 + \frac{4}{3} t^2 + 2t^2 + t^4 + 2t^3 + t^2 \right) \] Now, \( F(-1) \): - \( s^3 = -1 \), \( s^2 = 1 \), \( s = -1 \): \[ F(-1) = \left( -\frac{4}{3} t^4 - \frac{8}{3} t^3 + 2t^4 + 4t^3 -\frac{4}{3} t^2 + 2t^2 - t^4 - 2t^3 - t^2 \right) \] ### 4. Subtract \( F(-1) \) from \( F(1) \): \[ \text{Integral} = F(1) - F(-1) \] Let’s write them in a single line for each power of \( t \): #### For \( t^4 \): - \( F(1): \frac{4}{3}t^4 + 2t^4 + t^4 = \frac{4}{3}t^4 + 2t^4 + t^4 \) - \( F(-1): -\frac{4}{3}t^4 + 2t^4 - t^4 = -\frac{4}{3}t^4 + 2t^4 - t^4 \) - Difference: \( \left(\frac{4}{3} + 2 + 1\right) - \left(-\frac{4}{3} + 2 - 1\right) = \frac{4}{3} + 2 + 1 + \frac{4}{3} - 2 + 1 = \left(\frac{8}{3} + 2\right) = \frac{14}{3} t^4 \) #### For \( t^3 \): - \( F(1): \frac{8}{3}t^3 + 4t^3 + 2t^3 \) - \( F(-1): -\frac{8}{3}t^3 + 4t^3 - 2t^3 \) - Difference: \( \left(\frac{8}{3} + 4 + 2\right) - \left(-\frac{8}{3} + 4 - 2\right) = \frac{8}{3} + 4 + 2 + \frac{8}{3} - 4 + 2 = \frac{16}{3} + 2 = \frac{22}{3} t^3 \) #### For \( t^2 \): - \( F(1): \frac{4}{3}t^2 + 2t^2 + t^2 \) - \( F(-1): -\frac{4}{3}t^2 + 2t^2 - t^2 \) - Difference: \( \left(\frac{4}{3} + 2 + 1\right) - \left(-\frac{4}{3} + 2 - 1\right) = \frac{4}{3} + 2 + 1 + \frac{4}{3} - 2 + 1 = \frac{8}{3} + 2 = \frac{14}{3} t^2 \) ### **Final Answer:** \[ \boxed{ \int_{-1}^{1} \left( 4s^2 t^4 + 8s^2 t^3 + 4s t^4 + 8s t^3 + 4s^2 t^2 + 4s t^2 + t^4 + 2t^3 + t^2 \right) ds = \frac{14}{3} t^4 + \frac{22}{3} t^3 + \frac{14}{3} t^2 } \] This is the fully worked-out answer, integrated with respect to \( s \), evaluated from \( s = -1 \) to \( s = 1 \), **all together** as requested!