Let f(x)= ₀^t t (t^2-9 t+20 ) d t, 1 x 5 . If the range of f is [ , ] , then 4( + ) equals :

Step-by-step solution

f^ (x)=x (x^2-9 x+20 ), x (1,5)=(x-4) x(x-5) aligned & f^ (x)>0 x (1,4) & f^ (x) < 0 x (4,5) & f(x) increasing in (1,4) & f(x) decreasing in (4,5) & critical points to check: aligned aligned & x=1,4,5 & f(x)= ₀^x (t^3-9 t^2+20 t ) d t & = t^4 4 -3 t^3+ .10 t^2 |₀ ^x= x^4 4 -3 x^3+10 x^2 & f(1)= 1 4 -3+10= 29 4 & f(4)=4^3-3.4^3+10.4^2=-2.4^3+10.4^2=32 & f(5)= 5^4 4 -3.5^3+10.25= 5^4 4 -125= 125 4 aligned aligned & Range [ 29 4 , 32 ] 4( + )=128+29 & =157 aligned