1. Given function

We have

$$f(x)=\int_0^x t(t^2-9t+20)\,dt, \qquad 1\le x\le 5.$$

So the integrand is

$$t(t^2-9t+20)=t^3-9t^2+20t.$$

Thus

$$f(x)=\int_0^x (t^3-9t^2+20t)\,dt.$$

2. Find an explicit expression for $f(x)$

Integrating,

$$f(x)=\left[\frac{t^4}{4}-3t^3+10t^2\right]_0^x =\frac{x^4}{4}-3x^3+10x^2.$$

So

$$f(x)=\frac{x^4}{4}-3x^3+10x^2.$$

3. Find critical points in $[1,5]$

Differentiate:

$$f'(x)=x^3-9x^2+20x=x(x^2-9x+20)=x(x-4)(x-5).$$

Critical points are

$$x=0,4,5.$$

In the interval $[1,5]$, relevant critical points are $x=4$ and $x=5$.

4. Check monotonicity

For $1\le x<4$:

• $x>0$
• $x-4<0$
• $x-5<0$

Hence

$$f'(x)=x(x-4)(x-5)>0.$$

So $f$ is increasing on $[1,4]$.

For $4<x<5$:

• $x>0$
• $x-4>0$
• $x-5<0$

Hence

$$f'(x)<0.$$

So $f$ is decreasing on $[4,5]$.

Therefore:

• maximum occurs at $x=4$,
• minimum occurs at one of the endpoints $x=1$ or $x=5$.

5. Compute values at important points

At $x=1$:

$$f(1)=\frac{1}{4}-3+10=\frac{1}{4}+7=\frac{29}{4}.$$

At $x=4$:

$$f(4)=\frac{4^4}{4}-3(4^3)+10(4^2) =\frac{256}{4}-192+160 =64-192+160=32.$$

At $x=5$:

$$f(5)=\frac{5^4}{4}-3(5^3)+10(5^2) =\frac{625}{4}-375+250 =\frac{625}{4}-125 =\frac{125}{4}.$$

6. Find the range

Comparing the values:

$$f(1)=\frac{29}{4},\qquad f(4)=32,\qquad f(5)=\frac{125}{4}.$$

Since

$$\frac{29}{4}=7.25, \quad \frac{125}{4}=31.25, \quad 32=32,$$

we get

• minimum value $\alpha=\dfrac{29}{4}$,
• maximum value $\beta=32$.

So the range is

$$\left[\frac{29}{4},32\right].$$

7. Compute $4(\alpha+\beta)$

$$\alpha+\beta=\frac{29}{4}+32=\frac{29}{4}+\frac{128}{4}=\frac{157}{4}.$$

Therefore,

$$4(\alpha+\beta)=4\cdot \frac{157}{4}=157.$$

8. Match with options

Thus the correct option is

$$\boxed{157}$$

which is Option B.