Integral Conditions 7F0D61

1. **State the problem:** We want to find values of $a$ and $b$ such that for any linear function $g(x) = cx + d$, the integral $$\int_0^1 f(x)g(x) \, dx = 0$$ always holds. 2. **Express the integral:** Substitute $f(x) = ax^2 + bx + 1$ and $g(x) = cx + d$: $$\int_0^1 (ax^2 + bx + 1)(cx + d) \, dx = 0$$ 3. **Expand the integrand:** $$(ax^2 + bx + 1)(cx + d) = a c x^3 + a d x^2 + b c x^2 + b d x + c x + d$$ 4. **Rewrite integrand grouping powers of $x$:** $$a c x^3 + (a d + b c) x^2 + (b d + c) x + d$$ 5. **Integrate term-by-term from 0 to 1:** $$\int_0^1 a c x^3 \, dx = a c \frac{1}{4}$$ $$\int_0^1 (a d + b c) x^2 \, dx = (a d + b c) \frac{1}{3}$$ $$\int_0^1 (b d + c) x \, dx = (b d + c) \frac{1}{2}$$ $$\int_0^1 d \, dx = d$$ 6. **Sum the integrals:** $$a c \frac{1}{4} + (a d + b c) \frac{1}{3} + (b d + c) \frac{1}{2} + d = 0$$ 7. **Rewrite the equation grouping terms by $c$ and $d$:** $$c \left( a \frac{1}{4} + b \frac{1}{3} + \frac{1}{2} \right) + d \left( a \frac{1}{3} + b \frac{1}{2} + 1 \right) = 0$$ 8. **Since $c$ and $d$ are arbitrary (any linear $g(x)$), the coefficients must be zero:** $$a \frac{1}{4} + b \frac{1}{3} + \frac{1}{2} = 0$$ $$a \frac{1}{3} + b \frac{1}{2} + 1 = 0$$ 9. **Solve the system:** Multiply first equation by 12 to clear denominators: $$3a + 4b + 6 = 0$$ Multiply second equation by 6: $$2a + 3b + 6 = 0$$ 10. **Use elimination:** Multiply second equation by 3: $$6a + 9b + 18 = 0$$ Multiply first equation by 2: $$6a + 8b + 12 = 0$$ Subtract: $$(6a + 9b + 18) - (6a + 8b + 12) = 0 \Rightarrow b + 6 = 0 \Rightarrow b = -6$$ 11. **Find $a$:** Substitute $b = -6$ into $2a + 3b + 6 = 0$: $$2a + 3(-6) + 6 = 0 \Rightarrow 2a - 18 + 6 = 0 \Rightarrow 2a - 12 = 0 \Rightarrow 2a = 12 \Rightarrow a = 6$$ 12. **Final answer:** $$a = 6, \quad b = -6$$ These values ensure that $$\int_0^1 f(x)g(x)dx = 0$$ for any linear function $g(x)$.