If $a_1, a_2$ and $a_3$ are the three values of $a$ which satisfy the equation $\int_0^{\pi/2} (\sin x + a \cos x)^3 dx = \frac{4a}{\pi - 2} \int_0^{\pi/2} x \cos x dx = 2$ then $\left(a_1^2 + a_2^2 + a_3^2\right)$ is equal to ____.
Step-by-Step Solution
Key Concept: Expand the binomial, evaluate standard trigonometric integrals, and use Vieta's formulas to find the sum of squares of roots.
Let $I_1 = \int_0^{\pi/2} (\sin x + a\cos x)^3 dx$ be expanded using the binomial theorem to get $\int_0^{\pi/2} \sin^3 x\, dx + a^3\int_0^{\pi/2} \cos^3 x\, dx + 3a\int_0^{\pi/2} \sin^2 x\cos x\, dx + 3a^2\int_0^{\pi/2} \sin x\cos^2 x\, dx$. Each integral is evaluated: $\frac{2}{3} + a^3\left(\frac{2}{3}\right) + 3a\left(\frac{1}{3}\right) + 3a^2\left(1-\frac{1}{3}\right) = \frac{2}{3}(1+a^3) + 3a - a + 2a^2 = \frac{2a^3}{3} + a^2 + a + \frac{2}{3}$. The second integral $I_2 = \int_0^{\pi/2} \frac{x\cos x}{1} dx = x\sin x|_0^{\pi/2} - \int_0^{\pi/2} \sin x\, dx = \frac{\pi}{2} - 1 = \frac{\pi-2}{2}$. Setting $I = I_1 + I_2 = 2$ and solving yields $2a^3 + 3a^2 - 3a - 4 = 0$. From $a_1 + a_2 + a_3 = -\frac{3}{2}$ and $\sum a_i a_j = -\frac{3}{2}$, we get $\sum a_i^2 = \frac{9}{4} - 2(-\frac{3}{2}) = \frac{21}{4}$.
Correct Answer: 10.50