Key Concepts and Formulas

• Fundamental Theorem of Calculus (Part 1): If $F(x) = \int_a^x f(t) dt$, then $F'(x) = f(x)$. This theorem is crucial for finding the derivative of the given function $f(x)$.
• Finding the Range of a Continuous Function on a Closed Interval: For a continuous function $f(x)$ on a closed interval $[a, b]$, the absolute maximum and minimum values occur either at the critical points (where $f'(x) = 0$ or $f'(x)$ is undefined) within the interval or at the endpoints of the interval.
• Integration of Polynomials: The power rule for integration, $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (where $n \neq -1$), is used to evaluate the definite integral.

Step-by-Step Solution

Step 1: Understand the function and the goal. We are given a function $f(x)$ defined as a definite integral: $f(x)=\int\limits_0^x \mathrm{t}\left(\mathrm{t}^2-9 \mathrm{t}+20\right) \mathrm{dt}$ for $1 \leq x \leq 5$. We need to find the range of this function, denoted by $[\alpha, \beta]$, and then calculate $4(\alpha+\beta)$. The range $[\alpha, \beta]$ means $\alpha$ is the minimum value of $f(x)$ and $\beta$ is the maximum value of $f(x)$ on the interval $[1, 5]$.

Step 2: Find the derivative of $f(x)$ using the Fundamental Theorem of Calculus. According to the Fundamental Theorem of Calculus (Part 1), if $f(x) = \int_a^x g(t) dt$, then $f'(x) = g(x)$. In our case, $g(t) = t(t^2 - 9t + 20)$. Let's expand $g(t)$: $g(t) = t^3 - 9t^2 + 20t$ Therefore, the derivative of $f(x)$ is: $f'(x) = x(x^2 - 9x + 20) = x^3 - 9x^2 + 20x$

Step 3: Find the critical points of $f(x)$ by setting $f'(x) = 0$. Critical points are where the function's rate of change is zero or undefined. Since $f'(x)$ is a polynomial, it is defined everywhere. So, we only need to find where $f'(x) = 0$. $x^3 - 9x^2 + 20x = 0$ Factor out $x$: $x(x^2 - 9x + 20) = 0$ Now, factor the quadratic term: $x^2 - 9x + 20$. We need two numbers that multiply to 20 and add to -9. These numbers are -4 and -5. So, the equation becomes: $x(x - 4)(x - 5) = 0$ The solutions are $x = 0$, $x = 4$, and $x = 5$.

Step 4: Identify the critical points that lie within the given interval $[1, 5]$. The given interval is $1 \leq x \leq 5$. From the critical points found in Step 3 ($0, 4, 5$), the points that lie within $[1, 5]$ are $x = 4$ and $x = 5$. Note that $x=0$ is outside the interval.

Step 5: Evaluate $f(x)$ at the critical points within the interval and at the endpoints of the interval. The endpoints of the interval are $x=1$ and $x=5$. The critical points within the interval are $x=4$ and $x=5$. So we need to evaluate $f(x)$ at $x=1$, $x=4$, and $x=5$.

First, let's find the indefinite integral of $t(t^2 - 9t + 20) = t^3 - 9t^2 + 20t$. $\int (t^3 - 9t^2 + 20t) dt = \frac{t^4}{4} - \frac{9t^3}{3} + \frac{20t^2}{2} + C = \frac{t^4}{4} - 3t^3 + 10t^2 + C$

Now, let's evaluate the definite integral $f(x) = \int_0^x (t^3 - 9t^2 + 20t) dt$. $f(x) = \left[ \frac{t^4}{4} - 3t^3 + 10t^2 \right]_0^x$ $f(x) = \left( \frac{x^4}{4} - 3x^3 + 10x^2 \right) - \left( \frac{0^4}{4} - 3(0)^3 + 10(0)^2 \right)$ $f(x) = \frac{x^4}{4} - 3x^3 + 10x^2$

Now, evaluate $f(x)$ at $x=1$, $x=4$, and $x=5$:

• At $x=1$ (endpoint): $f(1) = \frac{1^4}{4} - 3(1)^3 + 10(1)^2 = \frac{1}{4} - 3 + 10 = \frac{1}{4} + 7 = \frac{1 + 28}{4} = \frac{29}{4}$

• At $x=4$ (critical point): $f(4) = \frac{4^4}{4} - 3(4)^3 + 10(4)^2 = 4^3 - 3(64) + 10(16) = 64 - 192 + 160 = 224 - 192 = 32$

• At $x=5$ (critical point and endpoint): $f(5) = \frac{5^4}{4} - 3(5)^3 + 10(5)^2 = \frac{625}{4} - 3(125) + 10(25) = \frac{625}{4} - 375 + 250 = \frac{625}{4} - 125 = \frac{625 - 500}{4} = \frac{125}{4}$

Step 6: Determine the minimum and maximum values of $f(x)$ on the interval $[1, 5]$. We have the following values: $f(1) = \frac{29}{4} = 7.25$ $f(4) = 32$ $f(5) = \frac{125}{4} = 31.25$

Comparing these values: The minimum value is $\frac{29}{4}$. The maximum value is $32$.

Therefore, the range of $f(x)$ on the interval $1 \leq x \leq 5$ is $[\alpha, \beta] = \left[\frac{29}{4}, 32\right]$. So, $\alpha = \frac{29}{4}$ and $\beta = 32$.

Step 7: Calculate $4(\alpha+\beta)$. We need to find $4(\alpha+\beta)$. $\alpha + \beta = \frac{29}{4} + 32 = \frac{29}{4} + \frac{32 \times 4}{4} = \frac{29}{4} + \frac{128}{4} = \frac{29 + 128}{4} = \frac{157}{4}$

Now, multiply by 4: $4(\alpha + \beta) = 4 \times \frac{157}{4} = 157$

Common Mistakes & Tips

• Forgetting to check endpoints: Always evaluate the function at the endpoints of the given interval, not just the critical points.
• Incorrectly factoring or solving the derivative: Double-check your algebra when solving $f'(x) = 0$. A small error can lead to the wrong critical points.
• Not considering the interval: Ensure that the critical points you identify are actually within the specified interval $[1, 5]$. Points outside this interval are irrelevant for determining the range on that specific interval.
• Calculation errors in evaluation: Be careful when substituting values into $f(x)$, especially with fractions and powers.

Summary

To find the range of the given function $f(x)$ on the interval $[1, 5]$, we first found its derivative $f'(x)$ using the Fundamental Theorem of Calculus. We then found the critical points by setting $f'(x) = 0$ and identified which of these critical points lay within the interval $[1, 5]$. Finally, we evaluated $f(x)$ at these critical points and at the endpoints of the interval. The smallest of these values gave us $\alpha$ (the minimum), and the largest gave us $\beta$ (the maximum). We then computed $4(\alpha+\beta)$.

The final answer is $\boxed{157}$.