Application of Integrals

e? F (x) = ∫ (3t²+2t+4F' (t)dt.
e? F (x)+e? F' (x) = 3x²+2x+4F' (x).
(e? -4)F' (x) = 3x²+2x-e? F (x).
F' (4) = (48+8-e? F (4)/ (e? -4).
Also F (3)=0, F (x)= (x³+x²-36)/ (e? -4) from solution. F (4)= (64+16-36)/ (e? -4) = 44/ (e? -4).
F' (4) = (56-e? (44/ (e? -4)/ (e? -4) = (56 (e? -4)-44e? )/ (e? -4)² = (12e? -224)/ (e? -4)².
α=12, β=4. α+β=16.

$A_1+A_2={\displaystyle {\int }_0^{\pi /2}cosxdx}$

$={\left(Sinx\right)}_0^{\pi /2}=1$
$A_1={\displaystyle {\int }_0^{\pi /4}\left(cosx-sinx\right)dx={\left(sinx+cosx\right)}_0^{\pi /4}}$

$=\frac{2}{\sqrt[]{2}}-1=\sqrt[]{2}-1$

$So{A}_2=1-\left(\sqrt[]{2}-1\right)=2-\sqrt[]{2}=\sqrt[]{2}\left(\sqrt[]{2}-1\right)$

$Now\frac{A_1}{A_2}=\frac{1}{\sqrt[]{2}}$             

  $P\left(\overline{A}\cap B\right)+P\left(A\cap \overline{B}\right)=1-k,$

$P\left(A\cap B\cap C\right)=k^2$             

$P\left(A\right)+P\left(B\right)-2P\left(A\cap B\right)=1-k$ .(i)

$P\left(B\right)+P\left(C\right)-2P\left(B\cap C\right)=1-k$ .(ii)

$P\left(A\right)+P\left(C\right)-2P\left(A\cap C\right)=1-2k$ .(iii)

Adding (i), (ii) and (iii) we get $P\left(A\cup B\cup C\right)=\frac{-4k+3}{2}+k^2$

$\Rightarrow P\left(A\cup B\cup C\right)=\frac{2k^2-4k+2+1}{2}=\frac{2{\left(k-1\right)}^2+1}{2}\Rightarrow P\left(A\cup B\cup C\right)>\frac{1}{2}$

$\frac{1}{2}{\displaystyle \underset{0}{\overset{3/2}{\int }}\left(1+4x-x^2\right)dx}$        

$=\frac{1}{2}{\left[x+2x^2-\frac{x^3}{3}\right]}_0^{3/2}=\frac{39}{8}$

${\displaystyle \underset{0}{\overset{3/2}{\int }}mx=\frac{9m}{8}}$              

(as per question)

$\Rightarrow \frac{39}{8}=\frac{9m}{4}$

$m=\frac{39}{18}$

12m = 26

$R_2={\displaystyle \underset{0}{\overset{1}{\int }}\left(x-x^3\right)dx={\left(\frac{x^2}{2}-\frac{x^4}{4}\right)}_0^1}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$

$R_1={\displaystyle \underset{0}{\overset{\frac{1}{4}}{\int }}\left(\sqrt[]{x}-2x\right)dx={\left(\frac{2x^{3/2}}{3}-x^2\right)}_0^{\frac{1}{4}}}=\frac{1}{12}-\frac{1}{16}$

$=\frac{4-3}{48}=\frac{1}{48}$

$R_1+R_2={{\displaystyle \underset{0}{\overset{1}{\int }}\left(\sqrt[]{x}-x^3\right)dx=\left(\frac{2x^{3/2}}{3}-\frac{x^4}{4}\right)}}_0^1=\frac{2}{3}-\frac{1}{4}=\frac{5}{12}$

$\frac{R_1+R_2}{R_1}=1+\frac{R_2}{R_1}=\frac{\frac{5}{12}}{\frac{1}{48}}=20$ $\therefore \frac{R_2}{R_1}=19$

 

$\overrightarrow{a}=\alpha \widehat{i}+2\widehat{j}-\widehat{k}\&\overrightarrow{b}=-2\widehat{i}+\alpha \widehat{j}+\widehat{k}$

$\left|\overrightarrow{a}*\overrightarrow{b}\right|=\sqrt[]{15\left(a^2+4\right)}$

$2{\left|\overrightarrow{a}\right|}^2+\left(\overrightarrow{a}.\overrightarrow{b}\right){\left|\overrightarrow{b}\right|}^2=?$

$\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta$

$cos\theta =\frac{-1}{\left(a^2+5\right)}$ $\therefore \left|sin\theta \right|=\frac{\sqrt[]{{\left(a^2+5\right)}^2-1}}{\left(a^2+5\right)}$

$\left|\overrightarrow{a}*\overrightarrow{b}\right|=\left|\stackrel{?}{a}\right|*\left|\overrightarrow{b}\right|*\left|sin\theta \right|$

$=\left(a^5+5\right)*\frac{\sqrt[]{{\left(a^2+5\right)}^2-1}}{\left(a^2+5\right)}=\sqrt[]{15\left(a^2+4\right)}$

${\left(a^2+5\right)}^2-1=15\left(a^2+4\right)$

$\Rightarrow \left(\alpha =\pm 3\right)$

$2{\left|\overrightarrow{a}\right|}^2+\left(\overrightarrow{a}.\overrightarrow{b}\right){\left|\overrightarrow{b}\right|}^2$

= $2\left(a^2+5\right)-\left(a^2+5\right)$

$\left(a^2+5\right)=14$

  $P_1:2x-y-52=0,P_2:3x-y+4z-7=0\text{?}\text{?}P$

Equation of plane passing through the line of intersection between planes p₁ = 0 & p₂ = 0 is

    $P:P_1+\lambda P_2$            

$\therefore P:8x-y+32-14=0$      

It passes through the point (1, 0, 2)

Area of $\Delta ABC$

= $\frac{1}{2}AB\cdot BC$

$=\frac{1}{2}\cdot \sqrt[]{2}\cdot 1$

$=\frac{1}{\sqrt[]{2}}$

now required area

$={\displaystyle \underset{0}{\overset{4}{\int }}\left(2\sqrt[]{2}-\sqrt[]{2}x\right)dx-\frac{1}{\sqrt[]{2}}}$

$=\frac{32\sqrt[]{2}}{3}=8\sqrt[]{2}-\frac{1}{\sqrt[]{2}}$

$=\frac{13\sqrt[]{2}}{6}$

 ${\displaystyle \underset{3}{\overset{x}{\int }}f\left(x\right)dx}={\left(\frac{f\left(x\right)}{x}\right)}^3\Rightarrow x^3{\displaystyle \underset{3}{\overset{x}{\int }}f\left(x\right)dx}=f^3\left(x\right),$ differentiating w.r.to x

$x^{3}f\left(x\right)+3x^2\frac{f^3\left(x\right)}{x^3}=3f^2\left(x\right)f\text{'}\left(x\right)\Rightarrow 3y^2\frac{dy}{dx}=x^{3}y=\frac{3y^3}{x}\Rightarrow 3xy\frac{dy}{dx}=x^4+3y^2$

After solving we get $y^2=\frac{x^4}{3}+c{x}^2$ also curve passes through (3, 3) c = -2

$\therefore y^2=\frac{x^4}{3}-2x^2$ which passes through $\left(\alpha ,6\sqrt[]{10}\right)$ $\therefore \frac{{\alpha }^4-6{\alpha }^2}{3}=360\Rightarrow \alpha =6$

Coefficient of x in ${\left(1+x\right)}^p{\left(1-x\right)}^q={-}^p{C_0}^q{C}_1{+}^p{C_1}^q{C}_0=-3\Rightarrow p-q=3$

Coefficient of x² in ${\left(1+x\right)}^p{\left(1-x\right)}^q{=}^p{C_0}^q{C}_2{-}^p{C_1}^q{C}_1{-}^p{C_2}^q{C}_0=-5$

$\Rightarrow \frac{q\left(q-1\right)}{2}-pq+\frac{p\left(p-1\right)}{2}=-5\Rightarrow \frac{q\left(q-1\right)}{2}-\left(q-3\right)q+\frac{\left(q-3\right)\left(q-4\right)}{2}=-5\Rightarrow q=11,p=8$

Coefficient of x³ in ${\left(1+x\right)}^8{\left(1-x\right)}^{11}={-}^{11}{C}_3{+}^8{C_1}^{11}{C}_2{-}^8{C_2}^{11}{C}_1{+}^8{C}_3=23$