For $\mathrm{m}, \mathrm{n}>0$, let $\alpha(\mathrm{m}, \mathrm{n})=\int_{0}^{2} \mathrm{t}^{\mathrm{m}}(1+3 \mathrm{t})^{\mathrm{n}} dt$. If $11 \alpha(10,6)+18 \alpha(11,5)=\mathrm{p}(14)^{6}$, then p is equal to

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Definite Integrals

For

m, n > 0

, let

α (m, n) = \int_{0}^{2} t^{m} (1 + 3 t)^{n} d t

. If

11 α (10, 6) + 18 α (11, 5) = p (14)^{6}

, then p is equal to

Answer: 32

Use integration by parts on

11 α (10, 6) = 11 \int_{0}^{2} t^{10} (1 + 3 t)^{6} d t

. Let

u = (1 + 3 t)^{6}

d v = t^{10} d t

. Then

d u = 6 (1 + 3 t)^{5} \cdot 3 d t

v = \frac{t ^{11}}{11}

. Thus

11 α (10, 6) = [t^{11} (1 + 3 t)^{6}]_{0}^{2} - 18 \int_{0}^{2} t^{11} (1 + 3 t)^{5} d t

. Therefore

11 α (10, 6) + 18 α (11, 5) = [t^{11} (1 + 3 t)^{6}]_{0}^{2} = 2^{11} \cdot 7^{6} = 2^{11} \cdot 7^{6} = 2^{5} \cdot (2 \cdot 7)^{6} = 32 \cdot (14)^{6}

. Thus

p = 32

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Integer answer

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