Let f(x)= _0^xt(t^2-9t+20)dt,1 x 5.

Question

Let $f(x)=\int _{0}^{x}t({t}^{2}-9t+20)dt,1\le x\le 5$. If the range of $f$ is $[\alpha ,\beta ]$, then $4(\alpha +\beta )$ equals :

Accepted Answer

$f'(x) = x(x-4)(x-5)$, so $f(x) = \dfrac{x^4}4-3x^3+10x^2$.

On $(1,4)$, $f'(x)>0$ (increasing); on $(4,5)$, $f'(x)<0$ (decreasing). So $x=4$ gives a local max.

$$f(1) = \frac14-3+10 = \frac{29}4,\qquad f(4) = 64-192+160 = 32,\qquad f(5)=\frac{125}4$$

Since $f(1)<f(5)<f(4)$, the range is $[\alpha,\beta]=\left[\dfrac{29}4,32\right]$.

$$4(\alpha+\beta) = 4\left(\frac{29}4+32\right) = 29+128 = 157$$

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