If \lim\limits _t arrow 0(\int\limits_0^1(3 x+5)^t d x)^\frac1t=\frac\alpha5 e(\frac85)^\frac23 , then \alpha is equal to \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2025 · 29 Jan · Shift 2 · Q49 with Solution

Let $I(t)=\int_0^1 (3x+5)^t\,dx.$ We need $\lim_{t\to 0} \big(I(t)\big)^{1/t}.$

Observe that when $t\to 0$ , $(3x+5)^t = e^{t\ln(3x+5)} = 1+t\ln(3x+5)+o(t).$ Hence,

=1+t\int_0^1 \ln(3x+5)\,dx+o(t).$$

Now use the standard limit: if $I(t)=1+At+o(t),$ then $\lim_{t\to 0} \big(I(t)\big)^{1/t}=e^A.$ Therefore,

=\exp\left(\int_0^1 \ln(3x+5)\,dx\right).$$

Compute the integral: $\int_0^1 \ln(3x+5)\,dx.$ Put $u=3x+5 \implies du=3dx, \quad dx=\frac{du}{3}.$ When $x=0$ , $u=5$ ; when $x=1$ , $u=8$ .

So, $\int_0^1 \ln(3x+5)\,dx=\frac13\int_5^8 \ln u\,du.$

Use $\int \ln u\,du=u\ln u-u.$ Thus,

=\frac13\left((8\ln 8-8)-(5\ln 5-5)\right).$$ Simplifying, $$=\frac13(8\ln 8-5\ln 5-3).$$

Therefore the limit is

=e^{-1}\cdot 8^{8/3}\cdot 5^{-5/3}.$$ Since $$8^{8/3}=(2^3)^{8/3}=2^8=256,$$ we get $$\text{Limit}=\frac{256}{5^{5/3}e}.$$

Rewrite in the given form: $\frac{\alpha}{5e}\left(\frac85\right)^{2/3}.$

Note that

=\frac{\alpha}{5e}\cdot \frac{8^{2/3}}{5^{2/3}} =\frac{\alpha}{5e}\cdot \frac{4}{5^{2/3}} =\frac{4\alpha}{5^{5/3}e}.$$ Compare with $$\frac{256}{5^{5/3}e}.$$ Hence, $$4\alpha=256 \implies \alpha=64.$$

So the required integer is $\boxed{64}.$