Key Concepts and Formulas

• Integration by Parts (IBP) for Definite Integrals: $\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du$. This formula is used to integrate a product of two functions.
• Product Rule for Differentiation: $\frac{d}{dx}(uv) = u'v + uv'$. Integration by Parts is derived from this rule.
• Fundamental Theorem of Calculus: $\int_a^b f'(x) \, dx = f(b) - f(a)$ and $\int_a^b f''(x) \, dx = f'(b) - f'(a)$.

Step-by-Step Solution

We are given a twice-differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ with $f(2)=1$. We are also given $F(x) = xf(x)$ for all $x \in \mathbb{R}$. We need to find the value of $F^{\prime}(2)+\int\limits_0^2 F(x) d x$.

Step 1: Evaluate the first given integral using Integration by Parts. We are given $\int\limits_0^2 x F^{\prime}(x) d x=6$. We apply Integration by Parts with $u=x$ and $dv=F'(x)dx$. Then, $du = dx$ and $v = \int F'(x)dx = F(x)$. Using the IBP formula: $\int\limits_0^2 x F^{\prime}(x) d x = [x F(x)]_0^2 - \int\limits_0^2 F(x) d x$ We know $F(x) = xf(x)$. To find $F(2)$, we substitute $x=2$: $F(2) = 2f(2)$. Since $f(2)=1$, $F(2) = 2(1) = 2$. Now, evaluate the term $[x F(x)]_0^2$: $[x F(x)]_0^2 = (2 \cdot F(2)) - (0 \cdot F(0)) = 2(2) - 0 = 4$ Substitute this back into the IBP equation: $6 = 4 - \int\limits_0^2 F(x) d x$ Solving for $\int\limits_0^2 F(x) d x$: $\int\limits_0^2 F(x) d x = 4 - 6 = -2 \quad \ldots (1)$

Step 2: Evaluate the second given integral using Integration by Parts. We are given $\int\limits_0^2 x^2 F^{\prime \prime}(x) d x=40$. We apply Integration by Parts with $u=x^2$ and $dv=F''(x)dx$. Then, $du = 2x \, dx$ and $v = \int F''(x)dx = F'(x)$. Using the IBP formula: $\int\limits_0^2 x^2 F^{\prime \prime}(x) d x = [x^2 F^{\prime}(x)]_0^2 - \int\limits_0^2 F^{\prime}(x) \cdot 2x \, d x$ Evaluate the term $[x^2 F^{\prime}(x)]_0^2$: $[x^2 F^{\prime}(x)]_0^2 = (2^2 \cdot F^{\prime}(2)) - (0^2 \cdot F^{\prime}(0)) = 4F^{\prime}(2)$ The integral term can be written as $2 \int\limits_0^2 x F^{\prime}(x) d x$. From Step 1, we know that $\int\limits_0^2 x F^{\prime}(x) d x = 6$. Substitute these values back into the equation: $40 = 4F^{\prime}(2) - 2(6)$ $40 = 4F^{\prime}(2) - 12$ Now, solve for $F^{\prime}(2)$: $4F^{\prime}(2) = 40 + 12$ $4F^{\prime}(2) = 52$ $F^{\prime}(2) = \frac{52}{4} = 13 \quad \ldots (2)$

Step 3: Compute the required expression. We need to find $F^{\prime}(2)+\int\limits_0^2 F(x) d x$. Using the results from equation (1) and equation (2): $F^{\prime}(2)+\int\limits_0^2 F(x) d x = 13 + (-2) = 11$

Common Mistakes & Tips

• Incorrect choice of $u$ and $dv$: Always choose $u$ such that its derivative $du$ simplifies the integral, and $dv$ such that its integral $v$ is easily obtainable. The ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential) can be a helpful guide for choosing $u$.
• Errors in evaluating the definite limits: Carefully substitute the upper and lower limits into the $[uv]_a^b$ term and subtract.
• Algebraic mistakes: Double-check all arithmetic and algebraic manipulations, especially when solving for unknown quantities.

Summary

The problem involves using Integration by Parts twice to evaluate two given definite integrals. The first integral provides a value for $\int\limits_0^2 F(x) d x$, and the second integral, along with the result from the first, allows us to determine the value of $F^{\prime}(2)$. Finally, these two values are summed to obtain the required expression.

The final answer is $\boxed{11}$.