Definite Integrals Part 2

f(x)=

x2

∫

0

‌

t2−5t+4

2+et

‌dt
f′(x)=2x(‌

x4−5x2+4

2+ex2

)=0
x=0, or (x2−4)(x2−1)=0
x=0,x=±2,±1
Now, f′(x)=‌

2x(x+1)(x−1)(x+2)(x−2)

(ex2+2)

f′(x) changes sign from positive to negative at x=−1,1 So, number of local maximum
points = 2
f′(x) changes sign from negative to positive at x=−2,0,2 So, number of local minimum
points = 3
∴m=2,n=3