Question

The value of c ∫[f(cx) + 1]dx – ∫f(c² + x)dx, for x ∈ [1 + c, a + c], [c, ac] c ≠ 0, is equal to........... 

(a) 0

(b) c(a – 1)

(c) ac

(d) a(c + 1)

Answer 1

The correct option (b) c(a – 1)  

Explanation:

I = c[^a+c∫_1+cf(cx)dx + ^a+c∫_1+cdx] – ^ac∫_cf(c² + x)dx 

put cx = c² + t 

hence 

when x = c + 1 

then t = c and 

when x = a + c 

then t = ac. 

∴ I = c[^ac∫_cf(c² + t)(dt/c)] + c[a + c – 1 – c] – ^ac∫_cf(c² + x)dx

= ^ac∫_cf(c² + t)dt – ^ac∫_cf(c² + x)dx + c(a – 1) 

= c(a – 1).