Question

$\int\limits^2_1 3ˣ dx,Obtain the given integrals as the limit of a sum.$

Answer 1

HELLO DEAR,


$\sf{\int\limits^2_1 3^x\,dx}$

we know if the integral is in the form of a^x then it's integral is a^x/loga


so, $\sf{\int\limits^2_1 3^x\,dx = [3^x/log3]^2_1}$

$\sf{[3^2/log3 - 3^1/log3]}$

$\sf{[9/log3 - 3/log3]}$

$\sf{6/log3}$


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

we have to obtain the given integral as the limits of a sum.
we know, if $\int\limits^b_a{f(x)}\,dx$ is given then, it equals to,
$=(b-a)\displaystyle\lim_{n\to \infty}\frac{1}{n}[f(a)+f(a+h)+f(a+2h)+......f\{a+(n-1)h\}]$
where h = (b - a)/n

Here, b = 2 and a = 1
so, h = (2 - 1)/n = 1/n

now, f(x) = 3^x
f(a) = f(1) = 3
f(a + h) = f(1 + h) = 3^(1 + h) = 3^(1 + 1/n) = 3.3^1/n
f(a + 2h) = f(1 + 2h) = 3^(1 + 2h) = 3^(1 + 2/n) = 3.3^2/n
.......................
..................................
f[a + (n - 1)h] = f[1 + (n -1)h] = 3^{1 + (n - 1)h} = 3^{1 + (n - 1)/n} = 3.3^(n-1)/n

so, f(1) + f(1 + h) + f(1 + 2h) + f(1 + 2h) + f(1 + 3h)......... + f[1 + (n - 1)h] = 3 + 3.3^1/n + 3.3^2/n + ....3.3^(n-1)/n
= 3[1 + 3^1/n + 3^2/n + 3^3/n +..... + 3^(n-1)/n]
= 3[(3^1/n)^n - 1]/(3^1/n - 1)]
= 6/(3^1/n - 1)

now, $\int\limits^2_1=(2-1)\lim_{n\to \infty}\frac{1}{n}\frac{6}{3^{1/n}-1}$

= $\lim_{n\to \infty}6\frac{\frac{1}{n}}{3^{1/n}-1}$

= $\frac{6}{ln3}$