Q.

If limt→0(∫01(3x+5)tdt)1t=α5e(85)23, then α is equal to _________.          [2025]


Ans.

(64)

Let L=limt→0(∫01(3x+5)tdx)1t

           =elimt→01t(∫01(3x+5)tdx–1)

           =elimt→01t[((3x+5)t+13(t+1))01–1]          [∵ 1∞ form]

            =elimt→0[8t+1–5t+1–3t–33t(t+1)]

            =elimt→0{[8·8t–5·5t–3t–(8–5)3t]·limt→0(1t+1)}

             =elimt→013[8(8t–1)t–5(5t–1)t–3tt]

              =e(8 ln(8)–5 ln(5)–3)3=e[ln(8)8/3–ln(5)5/3–ln e]

=(8)8/3(5)5/3·e ⇒ (85)23·825·1e=α5e(85)23          [Given]

On comparing, we get

∴  α=82=64.