Let f(x)=∫0xt(t2−9t+20)dt, 1≤x≤5. If the range of f is [α,β], then 4(α+β) equals:
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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)=∫axg(t)dt, then f′(x)=g(x). In this case, g(t)=t(t2−9t+20), so f′(x)=x(x2−9x+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 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.
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 t3−9t2+20t to get 4t4−3t3+10t2. 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)=429=7.25, f(4)=32, and f(5)=4125=31.25, we find that the minimum value is 429 and the maximum value is 32. Thus, α=429 and β=32.
Step 5: Calculate 4(α + β)
Finally, we substitute the values of α and β into the expression 4(α+β) and simplify to get the final answer.