To find the critical points, we set the derivative f'(x) equal to zero. We factor the quadratic expression x² - 9x + 20 into (x-4)(x-5). This gives us the critical points x=0, x=4, and x=5. We are interested in the interval 1 ≤ x ≤ 5.

To find the critical points, we set the derivative f'(x) equal to zero. We factor the quadratic expression x² - 9x + 20 into (x-4)(x-5). This gives us the critical points x=0, x=4, and x=5. We are interested in the interval 1 ≤ x ≤ 5.

We need to evaluate f(x) at the endpoints of the interval (x=1 and x=5) and at any critical points within the interval (x=4). We integrate t³ - 9t² + 20t to get (t^4)/(4) - 3t³ + 10t². Then we substitute the limits of integration to find the values of f(1), f(4), and f(5).

We need to evaluate f(x) at the endpoints of the interval (x=1 and x=5) and at any critical points within the interval (x=4). We integrate t³ - 9t² + 20t to get (t^4)/(4) - 3t³ + 10t². Then we substitute the limits of integration to find the values of f(1), f(4), and f(5).

The range of f(x) on the interval [1, 5] is given by the minimum and maximum values of f(x) at the endpoints and critical points. Comparing the values f(1) = (29)/(4) = 7.25, f(4) = 32, and f(5) = (125)/(4) = 31.25, we find that the minimum value is (29)/(4) and the maximum value is 32. Thus, α = (29)/(4) and β = 32.

The range of f(x) on the interval [1, 5] is given by the minimum and maximum values of f(x) at the endpoints and critical points. Comparing the values f(1) = (29)/(4) = 7.25, f(4) = 32, and f(5) = (125)/(4) = 31.25, we find that the minimum value is (29)/(4) and the maximum value is 32. Thus, α = (29)/(4) and β = 32.

Finally, we substitute the values of α and β into the expression 4(α + β) and simplify to get the final answer.

Let $f(x)=\int_0^x t\left(t^2-9t+20\right) dt$ , $1 \leq x \leq 5$ . If the range of $f$ is $[\alpha, \beta]$ , then $4(\alpha+\beta)$ equals:

Answer: 157

Step-by-step solution

Step 1: Find the derivative of f(x)

To find the critical points of $f(x)$ , we first need to find its derivative, $f'(x)$ . Using the Fundamental Theorem of Calculus, if $f(x) = \int_a^x g(t) dt$ , then $f'(x) = g(x)$ . In this case, $g(t) = t(t^2 - 9t + 20)$ , so $f'(x) = x(x^2 - 9x + 20)$ .

Step 2: Find critical points by setting f'(x) = 0

To find the critical points, we set the derivative $f'(x)$ equal to zero. We factor the quadratic expression $x^2 - 9x + 20$ into $(x-4)(x-5)$ . This gives us the critical points $x=0$ , $x=4$ , and $x=5$ . We are interested in the interval $1 \leq x \leq 5$ .

Step 3: Evaluate f(x) at critical points and endpoints

We need to evaluate $f(x)$ at the endpoints of the interval ( $x=1$ and $x=5$ ) and at any critical points within the interval ( $x=4$ ). We integrate $t^3 - 9t^2 + 20t$ to get $\frac{t^4}{4} - 3t^3 + 10t^2$ . Then we substitute the limits of integration to find the values of $f(1)$ , $f(4)$ , and $f(5)$ .

Step 4: Determine the range [α, β]

The range of $f(x)$ on the interval $[1, 5]$ is given by the minimum and maximum values of $f(x)$ at the endpoints and critical points. Comparing the values $f(1) = \frac{29}{4} = 7.25$ , $f(4) = 32$ , and $f(5) = \frac{125}{4} = 31.25$ , we find that the minimum value is $\frac{29}{4}$ and the maximum value is $32$ . Thus, $\alpha = \frac{29}{4}$ and $\beta = 32$ .

Step 5: Calculate 4(α + β)

Finally, we substitute the values of $\alpha$ and $\beta$ into the expression $4(\alpha + \beta)$ and simplify to get the final answer.

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Let f(x)=∫0xt(t2−9t+20)dtf(x)=\int_0^x t\left(t^2-9t+20\right) dtf(x)=∫0x​t(t2−9t+20)dt, 1≤x≤51 \leq x \leq 51≤x≤5. If the range of fff is [α,β][\alpha, \beta][α,β], then 4(α+β)4(\alpha+\beta)4(α+β) equals: