Test Index

Definite Integrals Part 3

Section: Mathematics

Question : 15 of 100

Marks: +1, -0

Let

f(x)=\int\limits_{0}^{x} t (t^2-9 t+20) \, dt, \;\; 1 \le x \le 5

t

f

[\alpha, \beta]

4(\alpha+\beta)

Solution: 👈: Video Solution

\;f'(x)=x (x^2-9 x+20) , x \in (1,5)

\;=(x-4) x(x-5)

\Rightarrow f'(x)>0 \forall x \in (1,4)

\Rightarrow f'(x)<0 \forall x \in (4,5)

\Rightarrow f(x) \;\text{ increasing in }\;(1,4)

\;f(x) \;\text{ decreasing in }\;(4,5)

⇒ critical points to check:

\;x=1,4,5

\;f(x)=\int\limits_{0}^{x} (t^3-9 t^2+20 t) \, dt

\;=\;\frac{t^4}{4}-3 t^3+10 t^2 \left.\right|_{0}^{x}=\;\frac{x^4}{4}-3 x^3+10 x^2

\;f(1)=\;\frac{1}{4}-3+10=\;\frac{29}{4}

\;f(4)=4^3-3\cdot4^3+10\cdot4^2=-2\cdot4^3+10\cdot4^2=32

\;f(5)=\;\frac{5^4}{4}-3\cdot5^3+10\cdot5^2=\;\frac{5^4}{4}-125=\;\frac{125}{4}

\;\;\text{ Range }\; \Rightarrow \left[\;\frac{29}{4}, 32\right] \Rightarrow 4(\alpha+\beta)=128+29

\;=157

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