Maths M.C.Q (1 Marks)

4. $\frac{1}{2}\text{x}|\text{x}|+\text{c}$

4. $\text{a}=\frac{1}{3},\text{ b}=-1$

Solution:

$\text{I}=\int\frac{\text{x}^3}{\sqrt{1+\text{x}^2}}\text{ dx}$

$1+\text{x}^2=\text{t}$

$2\text{xdx}=\text{dt}$

$\text{x dx}=\frac{\text{dt}}{2}$

$\text{I}=\int\frac{\text{x}^2}{\sqrt{1+\text{x}^2}}\text{x dx}$

$\text{I}=\int\frac{\text{t}-1}{\sqrt{\text{t}}}\frac{\text{dt}}{2}$

$\text{I}=\frac{1}{2}\Big(\frac{2}{3}\text{t}^{\frac{3}{2}}-2\sqrt{\text{t}}\Big)+\text{C}$

$\text{I}=\frac{1}{3}(1+\text{x}^{2})^{\frac{3}{2}}-\sqrt{1+\text{x}^2}+\text{C}$

$\text{a}=\frac{1}{3},\text{ b}=-1$

3. $-\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$

Solution:

$\text{I}=\int\frac{\sin\text{x}}{3+4\cos^2\text{x}}\text{ dx}$

Put $\cos\text{x}=\text{t}$

$-\sin\text{x dx}=\text{dt}$

$\sin\text{x dx}=-\text{dt}$

$\text{I}=\int\frac{-\text{dt}}{3+4\text{t}^2}$

$\text{I}=\frac{-1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\text{t}}{\sqrt{3}}\Big)+\text{C}$

$\text{I}=\frac{-1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$

4. $\text{a}=\frac{1}{3},\text{b}=-1$

2. $0$

Solution:

$\int\limits^1_0(\text{a}-\text{x})^2\text{ f}(\text{x})\text{dx}$

$=\text{a}^2\int\limits^1_0\text{f}(\text{x})\text{dx}+\int\limits^1_0\text{x}^2\text{f}(\text{x})\text{dx}-2\text{a}\int\limits^1_0\text{x}\text{f}(\text{x})\text{dx}$

$=\text{a}^2\times1+\text{a}^2-2\text{aa}$ (As per given values)

$=2\text{a}^2-2\text{a}^2$

$=0$

1. $(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{c}$

3. $\frac{-1}{3}\cos\text{x}^3+\text{c}$

3. $\frac{\pi}{12}$

Solution:

Let, $\text{I}=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{1+\sqrt{\cot\text{x}}}\text{dx}\ ...{\text{(i)}}$

$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{\sqrt{\cot\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}\text{dx}$ $\bigg[\text{using}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}=\int\limits^\text{b}_\text{a}\text{f}\big(\text{a}+\text{b}-\text{x}\big)\text{dx}\bigg]$

$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{1+\sqrt{\tan\text{x}}}\text{dx}\ ...(\text{ii})$

Adding (i) and (ii) we get

$2\text{I}=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\bigg[\frac{1}{1+\sqrt{\cot\text{x}}}+\frac{1}{1+\sqrt{\tan\text{x}}}\bigg]\text{dx}$

$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}{\big(1+\sqrt{\cot\text{x}}\big)+\big(1+\sqrt{\tan\text{x}}\big)}\text{ dx}$

$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\Bigg[\frac{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}{2+\sqrt{\cot\text{x}+\sqrt{\tan\text{x}}}}\Bigg]\text{dx}$

$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\text{dx}$

$=\big[\text{x}\big]^\frac{\pi}{3}_\frac{\pi}{6}$

$=\frac{\pi}{3}-\frac{\pi}{6}$

$=\frac{\pi}{6}$

Hence, $\text{I}=\frac{\pi}{12}$

3. $\frac{1}{16}$

Solution:

$\int\frac{\cos8\text{x}+1}{\tan2\text{x}-\cot2\text{x}}\text{ dx}$

$=\int\frac{2\cos^24\text{x}}{\frac{\sin2\text{x}}{\cos2\text{x}}-\frac{\cos2\text{x}}{\sin2\text{x}}}\text{ dx}$

$=\int\frac{2\cos^24\text{x}}{\sin^22\text{x}-\cos^22\text{x}}\times\sin2\text{x}\cos2\text{x dx}$

$=\int-\frac{\cos^24\text{x}\sin4\text{x}}{\cos4\text{x}}\text{ dx}$

$=\frac{-1}{2}\int\sin8\text{x dx}$

$=\frac{\cos8\text{x}}{16}+\text{C}$

$\text{a}=\frac{1}{16}$

1. $\frac{1}{3}\text{e}^{\text{x}^3}+\text{C}$

Let $\text{I}=\int\text{x}^2\text{e}^{\text{x}^3}\text{dx}$

Also, let x³ = t ⇒ 3x² dx = dt

$\Rightarrow\ \text{I}=\frac{1}{3}\int\text{e}^\text{t}\text{dt}$

$=\frac{1}{3}(\text{e}^\text{t})+\text{C}$

$=\frac{1}{3}\text{e}^{\text{x}^3}+\text{C}$

3. $\frac{1}{2}\log\tan\Big(\frac{\pi}{2}+\frac{\pi}{3}\Big)+\text{C}$

Solution:

$\text{I}=\int\frac{1}{\cos\text{x}+\sqrt{3}\sin\text{x}}\text{ dx}$

$\text{I}=\frac{1}{2}\int\frac{2}{\frac{\cos\text{x}}{2}+\frac{\sqrt{3}}{2}\sin\text{x}}\text{ dx}$

$\text{I}=\frac{1}{2}\int\frac{1}{\cos\big(\text{x}-\frac{\pi}{6}\big)}\text{ dx}$

$\text{I}=\frac{1}{2}\int\sec\Big(\text{x}-\frac{\pi}{6}\Big)\text{dx}$

$\text{I}=\frac{1}{2}\ln\Big|\tan\Big(\frac{\text{x}}{2}+\frac{\pi}{3}\Big)\Big|+\text{C}$

1. $\sin\text{x}+\text{C}$

Solution:

$\int\text{x}\sin\text{x dx}=-\text{x}\cos\text{x}+\text{a}$

$\text{I}=\int\text{x}\sin\text{x dx}$

$\text{I}=\text{x}\int\sin\text{x dx}-\int\Big(\frac{\text{dx}}{\text{dx}}\int\sin\text{x dx}\Big)\text{dx}$

$\text{I}=-\text{x}\cos\text{x}+\int\cos\text{x dx}$

$\text{I}=\text{x}\cos\text{x}+\sin\text{x}+\text{C}$

$\text{a}=\sin\text{x}+\text{C}$

1. $\frac{1}{\text{y}}$

Solution:

$\int\frac{-1}{{\text{y}}^2}\text{dy}$

$=-\int\text{y}^{-2}\text{dy}$

$=\text{y}^{-1}=\frac{1}{\text{y}}$

1. $\frac{1}{4}\text{ In }(4\text{x}+5)+\text{c}$

Solution:

$\int\frac{\text{dx}}{4\text{x}+5}=\frac{1}{4}\int\frac{\text{dx}}{\text{x}}$ where x = 4x + 5

$=\frac{1}{4}\text{ In }\text{x}+\text{c}_{1}=\frac{1}{4}\text{ In }(4\text{x}+5)+\text{c}_{2}$

3. $\text{cosec}(\text{b}-\text{a)}\log\Big|\frac{\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}\Big|+\text{C}$

Solution:

Let $\text{I}=\frac{\text{dx}}{\sin(\text{x}-​​​​\text{a})\sin(\text{x}-\text{b})}$

$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{b}-\text{a})}{\sin(\text{x}-​​​​\text{a})\sin(\text{x}-\text{b})}\text{dx}$

$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{x}-\text{a}-\text{x}+\text{b})}{\sin(\text{x}-​​​​\text{a})\sin(\text{x}-\text{b})}\text{dx}$

$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin\big\{(\text{x}-\text{a})-(\text{x}+\text{b})\big\}}{\sin(\text{x}-​​​​\text{a})\sin(\text{x}-\text{b})}\text{dx}$

$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{x}-\text{a})\cos(\text{x}-\text{b})-\cos(\text{x}-\text{a})\sin(\text{x}-\text{b})}{\sin(\text{x}-​​​​\text{a})\sin(\text{x}-\text{b})}\text{dx}$

$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{x}-\text{a})\cos(\text{x}-\text{b})}{\sin(\text{x}-​​​​\text{a})\sin(\text{x}-\text{b})}-\frac{\cos(\text{x}-\text{a})\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a} )\sin(\text{x}-\text{b})}\text{dx}$

$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\cos(\text{x}-\text{b})}{\sin(\text{x}-​​​​\text{a})}-\frac{\cos(\text{x}-\text{a})}{\sin(\text{x}-\text{a} )}\text{dx}$

$=\frac{1}{\sin(\text{b}-\text{a})}\big[\log\sin|(\text{x}-\text{b})|-\log|\sin(\text{x}-\text{b})|\big]+\text{C}$

$=\text{cosec}(\text{b}-\text{a})\log\Big|\frac{\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}\Big|+\text{C}$

3. $\tan\frac{\text{x}}{2}+\text{k}$

3. $\big(\ln\text{x}\big)^{-1}\text{x}(\text{x}-1)$

Solution:

$\text{f}'(\text{x})=\frac{1}{\log_\text{e}\text{x}^3}(3\text{x}^2)-\frac{1}{\log_\text{e}\text{x}^2}(2\text{x})$

$=\frac{3\text{x}^2}{3\ln\text{ x}}-\frac{2\text{x}}{2\ln\text{ x}}$

$=\frac{\text{x}^2}{\ln\text{ x}^{-1}}-\frac{\text{x}}{\ln\text{ x}}$

$=\frac{1}{\ln\text{ x}}\text{x}(\text{x}-1)$

$=\big(\ln\text{ x}\big)^{-1}\text{x}(\text{x}-1)$

2. $\log_{10}\text{e.x}\log_{\text{e}}\big(\frac{\text{x}}{\text{e}}\big)+\text{c}$

2. $\frac{\pi}{4}$

4. $\frac{\pi}{2}$

3. $\frac{2}{3}\text{x}^\frac{3}{2}+2\text{x}^\frac{1}{2}+\text{c}$

Solution:

$\int\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)\text{dx}$

$=\int\text{x}^\frac{1}{2}\text{dx}+\int\text{x}^-\frac{1}{2}\text{dx}$

we know that $\int\text{x}^\text{n}\text{dx}=\frac{\text{x}^\text{n}+1}{\text{n}+1}$

$=\frac{\text{x}^\frac{3}{2}}{\frac{3}{2}}+\frac{\text{x}^\frac{1}{2}}{\frac{1}{2}}+\text{c}$

$=\frac{2}{3}\text{x}^\frac{3}{2}+2\text{x}^\frac{1}{2}+\text{c}$

4.  none of these.

Solution:

$\int|\text{x}|^3\text{ dx}$

$|\text{x}|=\begin{cases}\text{x},\text{ x}\geq0\\-\text{x},\text{ x}<0\end{cases}$

Case I:

When $\text{x}\geq0$

$\therefore\ \int|\text{x}|^3\text{ dx}$

$=\int\text{x}^3\text{ dx}$

$=\frac{\text{x}^4}{4}+\text{C}$

Case II:

$\text{x}<0$

$\int|\text{x}|^3\text{ dx}$

$=-\int\text{x}^3\text{ dx}$

$=\frac{-\text{x}^4}{4}+\text{C}$ 

1. $\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{C}$

Solution:

$\int\text{e}^\text{x}\Big(\frac{1-\text{x}}{1+\text{x}^2}\Big)^2\text{dx}$

$=\int\text{e}^\text{x}\frac{1+\text{x}^2-2\text{x}}{(1+\text{x}^2)^2}\text{dx}$

$=\int\text{e}^\text{x}\Big[\frac{1}{(1+\text{x}^2)}-\frac{2\text{x}}{(1+\text{x}^2)^2}\Big]\text{dx}$

$=\int\text{e}^\text{x}[\text{f(x)}+\text{f}'(\text{x})]\text{dx},$ where $\text{f(x)}=\frac{1}{1+\text{x}^2}$

$=\text{e}^\text{x}\text{f(x)}+\text{C}=\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{C}$

2. 0 

$\text{Let I}=\int^{1}\limits_{0}\tan^{-1}\bigg(\frac{2\text{x}-1}{1+\text{x}-\text{x}^{2}}\bigg)\text{dx}$

$\Rightarrow\text{I}=\int^{1}\limits_{0}\tan^{-1}\bigg(\frac{\text{x}-(1-\text{x})}{1+\text{x}(1-\text{x})}\bigg)\text{dx}$

$\Rightarrow\text{I}=\int^{1}\limits_{0}\Big[\tan^{-1}\text{x}-\tan^{-1}(1-\text{x)}\Big]\text{dx}$

$\Rightarrow\text{I}=\int^{\text{1}}\limits_{0}\Big[\tan^{-1}(1-\text{x)}-\tan^{-1}(1-1+\text{x)}\Big]\text{dx}$

$\Rightarrow\text{I}=\int^{1}\limits_{0}\Big[\tan^{-1}(1-\text{x)}-\tan^{-1}\text{(x)}\Big]\text{dx}$

$\Rightarrow\text{I}=\int^{1}_{0}\Big[\tan^{-1}(1-\text{x)}-\tan^{-1}\text{(x)}\Big]\text{dx}$

Adding (1) and (2), we obtain

$\Rightarrow2\text{I}=\int^{1}\limits_{0}\Big(\tan^{-1}\text{x)}+\tan^{-1}(1-\text{x)}-\tan^{-1}\text{x}\Big)\text{dx}$

$\Rightarrow2\text{I}=0$

$\Rightarrow\text{I}=0$

4. $\frac{\pi}{12}$

$\int\frac{\text{dx}}{1+\text{x}^{2}}=\tan^{-1}\text{x}=\text{F}\text{(x)}$

By second fundamental theorem of calculus, we obtain

$\int\limits_{1}^{\sqrt{3}}\frac{\text{dx}}{1+\text{x}^{2}}=\text{F}(\sqrt{3})-\text{F}(1)$

$=\tan^{-1}\sqrt{3}-\tan^{-1}1$

$=\frac{\pi}{3}-\frac{\pi}{4}$

$=\frac{\pi}{12}$

$\ \text{put}\ 3\text{x}=\text{t}\Rightarrow3\text{dx}=\text{dt}$

$\therefore\int\frac{\text{dx}}{(2)^{2}+(3\text{x})^{2}}=\frac{1}{3}\int\frac{\text{dt}}{(2)^{2}+\text{t}^{2}}$

$=\frac{1}{3}\bigg[\frac{1}{2}\tan^{-1}\frac{t}{2}\bigg]$

$=\frac{1}{6}\tan^{-1}\bigg(\frac{3\text{x}}{2}\bigg)$

$=\text{F}\text{(x)}$

1. $\tan\text{x}-\cot\text{x+c}$

Solution:

$\int\sec^2\text{x}.\text{cosec}^2\text{xdx}$

$=\int\frac{{1}}{{\cos^2\text{x.}\sin^2\text{x}}}\text{dx}$

$=\int\frac{{{{\cos^2\text{x.}\sin^2\text{x}}}}}{{\cos^2\text{x.}\sin^2\text{x}}}\text{dx}$

$=\int\frac{{{{\cos^2\text{x.}}}}}{{\cos^2\text{x}\sin^2\text{x}}}+\frac{{{{\sin^2\text{x}}}}}{{\cos^2\text{x.}\sin^2\text{x}}}\text{dx}$

$=\int(\text{cosec}^2\text{x}+\sec^2\text{x})\text{dx}$

$=-\cot\text{x}+\tan\text{x}+\text{c}$

$=\tan\text{x}-\cot\text{x+c}$

2. $\text{f}(\text{x})=\tan\text{x}$

Solution:

$\int\sin \text{xd}(\sec\text{x})=\int\sin\text{x}\sec\text{x}\tan\text{xdx}$

$=\int\tan^2\text{xdx}$

$=\int(\sec^2\text{x - 1})\text{dx}$

$=\tan\text{x - x}+\text{c}$

$\Rightarrow\text{f}(\text{x})=\tan\text{x},\text{g}(\text{x})=\text{x}$

$=\frac{2}{3}\text{x}^\frac{3}{2}+2\text{x}^\frac{1}{2}+\text{C}$

4. $4\tan\text{x}+9\cot\text{x}+25\text{x}+\text{c}$

1. $2\sin\text{x}-\ell\text{ n }\mid\sec\text{x}+\tan\text{x}\mid+\text{ c}$

4.  $\text{x}-\frac{\text{x}^2}{2}+\frac{\text{x}^3}{3}-\log|1+\text{x}|+\text{C}$

Solution:

Let $\text{I}=\int\frac{\text{x}}{\text{x}+1}\text{dx}$

We know that, $\frac{\text{x}^3}{\text{x}+1}$ is an improper fraction.

To convert it into proper fraction, we have to divide numerator by denominator.

After performing long division, we get

$\frac{\text{x}^3}{\text{x}+1}=(\text{x}^2-\text{x}+1)-\frac{1}{(\text{x}+1)}$

$\therefore\ \text{I}=\int\Big((\text{x}^2-\text{x}+1)-\frac{1}{(\text{x}+1)}\Big)\text{dx}$

$=\frac{\text{x}^3}{3}-\frac{\text{x}^2}{2}+\text{x}-\log|\text{x}+1|+\text{C}$ 

3. $\frac{\pi}{2}$

Solution:

We have,

$\text{I}=\int\limits^1_0\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big\}\text{dx}$

We know since $\int\text{f}'(\text{x})=\text{f}(\text{x})$

$\text{f}(\text{x})=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ and $\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big\}$

Therefore, $\text{I}=\Big[\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big]^1_0$

$=\sin^{-1}(1)-\sin^{-1}(0)$

$=\frac{\pi}{2}$

1. $\frac{1}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$

Solution:

$\text{I}=\int\text{x}\sec\text{x}^2\text{ dx}$

Put $\text{x}^2=\text{t}$

$=\text{x}=\sqrt{\text{t}}$

$2\text{xdx}=\text{dt}$

$\text{xdx}=\frac{\text{dt}}{2}$

$\text{I}=\int\sec\text{t}\frac{\text{dt}}{2}$

$\text{I}=\frac{1}{2}\log(\sec\text{t}+\tan\text{t})+\text{C}$

$\frac{1}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$

4. $\sin^{-1}(2\text{x})+\text{c}$

4. $\log(1+\log\text{x})$

Solution:

$\text{I}=\int\frac{1}{\text{x}+\text{x}\log\text{x}}\text{ dx}$

$\text{I}=\int\frac{\text{dx}}{\text{x}+(1+\log\text{x})}$

Put $1+\log\text{x}=\text{t}$

$\frac{1}{\text{x}}\text{dx}=\text{dt}$

$\text{I}=\int\frac{1}{\text{t}}\text{ dt}$

$\text{I}=\log|\text{t}|+\text{C}$

$\text{I}=\log(1+\log\text{x})+\text{C}$

3. 0

Solution:

$\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x }\log\tan\text{x dx}\ ....(\text{i})$

$\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin(\pi-2\text{x})\log\tan\big(\frac{\pi}{2}-\text{x}\big)\text{dx}$

$\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x}\log\cot\text{x dx}\ ...(\text{ii})$

Adding (i) and (ii) we get

$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x}\big(\log\tan\text{x}+\log\cot\text{x}\big)\text{dx}$

$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x}\big(\log\tan\text{x}\cot\text{x}\big)\text{dx}$

$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin\text{x}(\log1)\text{dx}$

$\text{I}=0$

4. $\frac{1}{10\text{x}}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{c}$

3. $\text{a}=-\frac{1}{10},\text{ b}=\frac{2}{5}$

Solution:

$\text{I}=\int\frac{1}{(\text{x}+2)(\text{x}^2+1)}\text{ dx}$

Consider,

$\frac{1}{(\text{x}+2)(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+2}+\frac{\text{Bx}+\text{C}}{(\text{x}^2+1)}$

$1=\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x}+2)$

Comaring coefficeints and solving it simultaneously we get

$\text{A}=\frac{1}{5},\text{ B}=-\frac{1}{5},\text{ C}=\frac{2}{5}$

$\text{I}=\int\bigg(\frac{1}{5\text{x}+1}+\frac{\frac{-1}{5}\text{x}+\frac{2}{5}}{\text{x}^2+1}\bigg)\text{dx}$

Integrating we get as,

$\frac{1}{5}\log|\text{x}+2|-\frac{1}{10}\log|\text{x}^2+1|+\frac{2}{5}\tan^{-1}\text{x}+\text{C}$

$\text{a}=-\frac{1}{10},\text{ b}=\frac{2}{5}$

3. $\cos\Big(\frac{\text{x}}{7}+10\Big)\text{C}$

Solution:

Let $\text{I}=\int\frac{1}{7}\sin\Big(\frac{\text{x}}{7}+10\Big)\text{dx}$

$=\frac{1}{7}\int\sin\Big(\frac{\text{x}}{7}+10\Big)=\frac{1}{7}\frac{-\cos\Big(\frac{\text{x}}{7}+10\Big)}{\frac{1}{7}}$

$=-\cos\Big(\frac{\text{x}}{7}+10\Big)+\text{C}$