Given-2-4-ax-n-1-dx-58-0-2-ax-n-1-dx-10-find-the-value-of-a-and-n-

Question Number 209557 by Spillover last updated on 14/Jul/24

G i v e n

\int_{2}^{4} ({a x}^{n} + 1) d x = 58 \int_{0}^{2} ({a x}^{n} + 1) d x = 10

f i n d t h e v a l u e o f a a n d n

Answered by mr W last updated on 14/Jul/24

\int_{0}^{2} ({a x}^{n} + 1) d x = {[\frac{{a x}^{n + 1}}{n + 1} + x]}_{0}^{2} = \frac{a 2^{n + 1}}{n + 1} + 2 = 10

\Rightarrow \frac{a 2^{n + 1}}{n + 1} = 8 \dots (i)

\int_{0}^{4} ({a x}^{n} + 1) d x = \frac{a 4^{n + 1}}{n + 1} + 4 = 10 + 58 = 68

\Rightarrow \frac{a 4^{n + 1}}{n + 1} = 64 \dots (i i)

(i i) / (i) : {(\frac{4}{2})}^{n + 1} = 8 \Rightarrow n = 2 ✓

\Rightarrow a = \frac{8 \times 3}{2^{3}} = 3 ✓

Commented by Spillover last updated on 14/Jul/24

t h a n k y o u

Answered by lepuissantcedricjunior last updated on 16/Jul/24

\int_{2}^{4} ({a x}^{n} + 1) d x = 58

\int_{0}^{2} ({a x}^{n} + 1) d x = 10

=> {\begin{cases} \int_{2}^{4} ({a x}^{n} + 1) d x \\ \int_{0}^{2} ({a x}^{n} + 1) d x \end{cases} => {\begin{cases} = {[\frac{a}{1 + n} x^{n + 1} + x]}_{2}^{4} \\ = {[\frac{a}{n + 1} x^{n + 1} + x]}_{0}^{2} \end{cases} => {\begin{cases} = \frac{2^{2 n + 2} a}{1 + n} - \frac{2^{n + 1} a}{n + 1} + 2 \\ = \frac{2^{n + 1} a}{n + 1} + 2 \end{cases}

=> {\begin{cases} \frac{2^{n + 1} a (2^{n + 1} - 1)}{n + 1} = 56 (1) \\ \frac{2^{n + 1} a}{n + 1} = 8 (2) r e m p l a c o n s d a n s (1) \end{cases}

=> 8 (2^{n + 1} - 1) = 56 => 2^{n + 1} = 8 = 2^{3}

=> n + 1 = 3 => n = 2

d e p l u s \frac{2^{n + 1} a}{n + 1} = 8 => \frac{8}{3} a = 8 => a = 3

d^{'} {o u}^{'} \Rightarrow {(n; a)} = {(2; 3)}

\dots \dots \dots \dots l e p u i s s a n t c e d r i c j u n i o r \dots \dots . .

Commented by Spillover last updated on 30/Jul/24

g r e a t