Question

Integrate 1/9+2x limit is 0-8

Answer 1

Answer:

$\mathrm{\dfrac{1}{2}}\mathrm{(ln3)}}$

Step-by-step explanation:

Given :

$\displaystyle \int \limits^8_0 \mathrm{\dfrac{1}{9 + 2x}\,dx}$

To find :

Evaluate the given interval

Solution :

$\longmapsto \displaystyle \int \limits^8_0 \mathrm{\dfrac{1}{9 + 2x}\,dx}$

Let us substitute t in place of 2x ; Let t = 2x

differentiating with respect to x, we get,

$\implies \mathrm{\dfrac{dt}{dx} = 2}$

$\implies \mathrm{dt = 2\, dx}$

$\implies \mathrm{dx = \dfrac{dt}{2}}$

Now limits also change to,

• If x = 0, then t = 2(0) = 0 => t = 0
• If x = 8, then t = 2(8) = 16 => t = 16

So, substituting all these in the interval

$\longmapsto \displaystyle \int \limits^{16}_0 \mathrm{\dfrac{1}{9 + t}\,\dfrac{dt}{2}}$

$\longmapsto \mathrm{\dfrac{1}{2}}\displaystyle \int \limits^{16}_0 \mathrm{\dfrac{1}{9 + t}\,dt}$

now, let, 9 + t = s

Differentiating with respect to s, we get,

$\implies \mathrm{\dfrac{d}{ds}(9 + t) = (1)}$

$\implies \mathrm{dt = ds}$

Now limits also change to,

• If t = 0, then t = 9 + 0 = 9 => t = 9
• If t = 16, then t = 9 + 16 = 27 => t = 27

$\longmapsto \mathrm{\dfrac{1}{2}}\displaystyle \int \limits^{27}_9 \mathrm{\dfrac{1}{s}\,ds}$

$\boxed{ \displaystyle \int \mathrm{\dfrac{1}{x}\, dx = ln|x|+c}}$

$\longmapsto \mathrm{\dfrac{1}{2}}\mathrm{[ln|s|]^{27}_9}}$

$\longmapsto \mathrm{\dfrac{1}{2}}\mathrm{(ln|27| - ln|9|)}}$

We know that,

$\boxed{\mathrm{lnb - lna = ln\bigg(\dfrac{b}{a}\bigg)}}$

So,

$\longmapsto \mathrm{\dfrac{1}{2}}\mathrm{ln\bigg(\dfrac{\not{27}^3}{\not9}\bigg)}}$

$\longmapsto \mathrm{\dfrac{1}{2}}\mathrm{(ln3)}}$

$\therefore \underline{\boxed{\displaystyle \int \limits^8_0 \mathbf{\dfrac{1}{9 + 2x}\,dx = \dfrac{1}{2}ln3}}}$

❖ Extra information

⟡ Some basic Integrals :

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx+c \\\\ \sf x^n \ (n \neq -1)& \sf \dfrac{x^{n+1}}{n+1} + c \\\\ \sf \dfrac{1}{x} & \sf logx+ c\\\\ \sf {e}^{x} & \sf {e}^{x}+c\\\\ \sf sinx & \sf - \: cosx+ c \\\\ \sf cosx & \sf \: sinx + c\\\\ \sf {sec}^{2} x & \sf tanx + c\\\\ \sf {cosec}^{2}x & \sf - cotx+ c \\\\ \sf secx \: tanx & \sf secx + c\\\\ \sf cosecx \: cotx& \sf -\: cosecx + c\end{array}} \\ \end{gathered}$

Hope it helps!!