Application of Integrals Class 12 Maths NCERT Solutions

Step-by-step solution

1. 1The parabola opens rightward with vertex at the origin; the vertical line x = 3 closes the region.
2. 2By symmetry about the x-axis, total area = 2 × area in the first quadrant.
3. 3In Q1, $y=2\sqrt{x}$
4. 4Area = $2{\int }_0^{3}2\sqrt{x}dx=4{\int }_0^3{x}^{1/2}dx=4{\left[\frac{2}{3}{x}^{3/2}\right]}_0^3$
5. 5= $\frac{8}{3}\cdot 3^{3/2}=\frac{8}{3}\cdot 3\sqrt{3}=8\sqrt{3}$

Step-by-step solution

1. 1In the first quadrant, $y=3\sqrt{x}$
2. 2Required area = ${\int }_2^{4}3\sqrt{x}dx=3{\left[\frac{2}{3}{x}^{3/2}\right]}_2^4$
3. 3= $2(4^{3/2}-2^{3/2})=2(8-2\sqrt{2})=16-4\sqrt{2}$

Step-by-step solution

1. 1In the first quadrant, $x=2\sqrt{y}$
2. 2Area = ${\int }_2^{4}2\sqrt{y}dy=2{\left[\frac{2}{3}{y}^{3/2}\right]}_2^4$
3. 3= $\frac{4}{3}(4^{3/2}-2^{3/2})=\frac{4}{3}(8-2\sqrt{2})=\frac{32-8\sqrt{2}}{3}$

Step-by-step solution

1. 1This is a standard ellipse with $a=4$ and $b=3$
2. 2By symmetry, total area = 4 × area in the first quadrant.
3. 3In Q1, $y=\frac{3}{4}\sqrt{16-x^2}$
4. 4Area = $4{\int }_0^4\frac{3}{4}\sqrt{16-x^2}dx=3{\int }_0^4\sqrt{16-x^2}dx$
5. 5Use ${\int }_0^a\sqrt{a^2-x^2}dx=\frac{\pi a^2}{4}$: Area = $3\cdot \frac{\pi \cdot 16}{4}=12\pi$
6. 6(Equivalently, the ellipse area formula $\pi ab=\pi \cdot 4\cdot 3=12\pi$.

Step-by-step solution

1. 1Here $a=2$ and $b=3$
2. 2Area of an ellipse is $\pi ab$
3. 3Area = $\pi \cdot 2\cdot 3=6\pi$

Step-by-step solution

1. 1sin x is positive on (0, π) and negative on (π, 2π), so take absolute values.
2. 2Area = ${\int }_0^{2\pi }|\sin x|dx={\int }_0^{\pi }\sin xdx+\left|{\int }_{\pi }^{2\pi }\sin xdx\right|$
3. 3= $[-\cos x{]}_0^{\pi }+\left|[-\cos x{]}_{\pi }^{2\pi }\right|=(1+1)+|(-1)-1|$
4. 4= 2 + 2 = 4

Step-by-step solution

1. 1In the first quadrant, $y=2\sqrt{a}\sqrt{x}$
2. 2By symmetry, area = 2 × area in Q1.
3. 3Area = $2{\int }_0^{a}2\sqrt{a}\sqrt{x}dx=4\sqrt{a}{\left[\frac{2}{3}{x}^{3/2}\right]}_0^a$
4. 4= $\frac{8\sqrt{a}}{3}\cdot a^{3/2}=\frac{8a^2}{3}$

Step-by-step solution

1. 1The line $y=\frac{x}{\sqrt{3}}$ and the circle $x^2+y^2=4$ meet where $x^2+\frac{x^2}{3}=4\Rightarrow x=\sqrt{3}$, i.e. at (√3, 1).
2. 2On [0, √3] the top boundary is the line; on [√3, 2] it is the circle arc $y=\sqrt{4-x^2}$
3. 3Area = ${\int }_0^{\sqrt{3}}\frac{x}{\sqrt{3}}dx+{\int }_{\sqrt{3}}^2\sqrt{4-x^2}dx$
4. 4First integral: ${\left[\frac{x^2}{2\sqrt{3}}\right]}_0^{\sqrt{3}}=\frac{\sqrt{3}}{2}$
5. 5Second integral: ${\left[\frac{x}{2}\sqrt{4-x^2}+2{\sin }^{-1}\frac{x}{2}\right]}_{\sqrt{3}}^2=\left(0+\pi \right)-\left(\frac{\sqrt{3}}{2}+\frac{2\pi }{3}\right)=\frac{\pi }{3}-\frac{\sqrt{3}}{2}$
6. 6Area = $\frac{\sqrt{3}}{2}+\frac{\pi }{3}-\frac{\sqrt{3}}{2}=\frac{\pi }{3}$

Step-by-step solution

1. 1Name the vertices A(−1,0), B(1,3), C(3,2).
2. 2AB: $y=\frac{3}{2}(x+1)$
3. 3BC: $y=\frac{7-x}{2}$
4. 4AC: $y=\frac{x+1}{2}$
5. 5Area = ${\int }_{-1}^1\left(\frac{3(x+1)}{2}-\frac{x+1}{2}\right)dx+{\int }_1^3\left(\frac{7-x}{2}-\frac{x+1}{2}\right)dx$
6. 6= ${\int }_{-1}^1(x+1)dx+{\int }_1^3(3-x)dx={\left[\frac{x^2}{2}+x\right]}_{-1}^1+{\left[3x-\frac{x^2}{2}\right]}_1^3$
7. 7= $\left(\frac{1}{2}+1\right)-\left(\frac{1}{2}-1\right)+\left(9-\frac{9}{2}\right)-\left(3-\frac{1}{2}\right)=2+2=4$

Step-by-step solution

1. 1x² + 1 ≥ 1 > 0 on [0, 3], so the curve stays above the x-axis.
2. 2Area = ${\int }_0^3(x^2+1)dx={\left[\frac{x^3}{3}+x\right]}_0^3=(9+3)-0=12$

Step-by-step solution

1. 1sin x ≥ 0 on [0, π].
2. 2Area = ${\int }_0^{\pi }\sin xdx=[-\cos x{]}_0^{\pi }=-\cos \pi +\cos 0=1+1=2$

Step-by-step solution

1. 1Both circles are centred at the origin with radii 2 and 3 respectively.
2. 2The required region is the ring between the two circles.
3. 3Area = $\pi (3^2)-\pi (2^2)=9\pi -4\pi =5\pi$

Step-by-step solution

1. 1x³ is negative on (−2, 0) and positive on (0, 2), so split the integral.
2. 2Area = $\left|{\int }_{-2}^0{x}^{3}dx\right|+{\int }_0^2{x}^{3}dx=\left|{\left[\frac{x^4}{4}\right]}_{-2}^0\right|+{\left[\frac{x^4}{4}\right]}_0^2$
3. 3= |0 − 4| + (4 − 0) = 4 + 4 = 8

Step-by-step solution

1. 1Circle: $x^2+y^2=\frac{9}{4}$, radius $\frac{3}{2}$. Parabola: $y=\frac{x^2}{4}$
2. 2Intersections: $x^2+\frac{x^4}{16}=\frac{9}{4}\Rightarrow x^4+16x^2-36=0\Rightarrow x^2=2\Rightarrow x=\pm \sqrt{2}$
3. 3On the first quadrant piece, the top curve is the circle, the bottom curve is the parabola.
4. 4Area = $2{\int }_0^{\sqrt{2}}\left(\sqrt{\frac{9}{4}-x^2}-\frac{x^2}{4}\right)dx$
5. 5Using $\int \sqrt{a^2-u^2}du=\frac{u}{2}\sqrt{a^2-u^2}+\frac{a^2}{2}{\sin }^{-1}\frac{u}{a}$ with a = 3/2:
6. 6First integral = ${\left[\frac{x}{2}\sqrt{\frac{9}{4}-x^2}+\frac{9}{8}{\sin }^{-1}\frac{2x}{3}\right]}_0^{\sqrt{2}}=\frac{\sqrt{2}}{4}+\frac{9}{8}{\sin }^{-1}\frac{2\sqrt{2}}{3}$
7. 7Second integral = ${\left[\frac{x^3}{12}\right]}_0^{\sqrt{2}}=\frac{\sqrt{2}}{6}$
8. 8Area = $2\left(\frac{\sqrt{2}}{4}+\frac{9}{8}{\sin }^{-1}\frac{2\sqrt{2}}{3}-\frac{\sqrt{2}}{6}\right)=\frac{\sqrt{2}}{6}+\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3}$

Step-by-step solution

1. 1The line is $x=2y+3$. It cuts the x-axis at (3, 0).
2. 2Intersection with the parabola: $y=\sqrt{2y+3}\Rightarrow y^2=2y+3\Rightarrow y=3$, giving (9, 3).
3. 3The bounded region runs along the parabola from (0, 0) to (9, 3), then down the line to (3, 0), then back along the x-axis.
4. 4Area = ${\int }_0^3\sqrt{x}dx+{\int }_3^9\left(\sqrt{x}-\frac{x-3}{2}\right)dx$
5. 5= $2\sqrt{3}+{\left[\frac{2}{3}{x}^{3/2}-\frac{(x-3{)}^2}{4}\right]}_3^9$
6. 6= $2\sqrt{3}+(18-9)-\left(2\sqrt{3}-0\right)=9$

Step-by-step solution

1. 1Intersections: $(mx{)}^2=4ax\Rightarrow m^{2}x\cdot x=4ax\Rightarrow x=0$ or $x=\frac{4a}{m^2}$
2. 2In Q1, the parabola gives $y=2\sqrt{ax}$ and the line is $y=mx$
3. 3Area = ${\int }_0^{4a/m^2}\left(2\sqrt{ax}-mx\right)dx={\left[\frac{4\sqrt{a}}{3}{x}^{3/2}-\frac{m{x}^2}{2}\right]}_0^{4a/m^2}$
4. 4= $\frac{4\sqrt{a}}{3}{\left(\frac{4a}{m^2}\right)}^{3/2}-\frac{m}{2}\cdot \frac{16a^2}{m^4}=\frac{32a^2}{3m^3}-\frac{8a^2}{m^3}=\frac{8a^2}{3m^3}$

Step-by-step solution

1. 1They intersect at (0, 0) and at (4a, 4a).
2. 2The parabola y² = 4ax gives $y=2\sqrt{ax}$; the other is $y=\frac{x^2}{4a}$
3. 3On (0, 4a), the first parabola lies above the second.
4. 4Area = ${\int }_0^{4a}\left(2\sqrt{ax}-\frac{x^2}{4a}\right)dx={\left[\frac{4\sqrt{a}}{3}{x}^{3/2}-\frac{x^3}{12a}\right]}_0^{4a}$
5. 5= $\frac{4\sqrt{a}}{3}(8a^{3/2})-\frac{64a^3}{12a}=\frac{32a^2}{3}-\frac{16a^2}{3}=\frac{16a^2}{3}$

Step-by-step solution

1. 1Centres (0, 0) and (2, 0), each of radius 2 — the circles cut at two points.
2. 2Intersections: $x^2+y^2=4$ and $x^2-4x+y^2=0\Rightarrow 4=4x\Rightarrow x=1$, so points $(1,\pm \sqrt{3})$
3. 3By symmetry the common (lens-shaped) area = 2 × the first quadrant piece.
4. 4In Q1, the upper boundary of the lens is the arc of circle 2, $y=\sqrt{4-(x-2{)}^2}$, on [0, 1], then the arc of circle 1, $y=\sqrt{4-x^2}$, on [1, 2].
5. 5Half area = ${\int }_0^1\sqrt{4-(x-2{)}^2}dx+{\int }_1^2\sqrt{4-x^2}dx$
6. 6First integral (u = x − 2): ${\int }_{-2}^{-1}\sqrt{4-u^2}du={\left[\frac{u}{2}\sqrt{4-u^2}+2{\sin }^{-1}\frac{u}{2}\right]}_{-2}^{-1}=\frac{2\pi }{3}-\frac{\sqrt{3}}{2}$
7. 7Second integral: ${\left[\frac{x}{2}\sqrt{4-x^2}+2{\sin }^{-1}\frac{x}{2}\right]}_1^2=\pi -\left(\frac{\sqrt{3}}{2}+\frac{\pi }{3}\right)=\frac{2\pi }{3}-\frac{\sqrt{3}}{2}$
8. 8Half area = $\frac{4\pi }{3}-\sqrt{3}$; total = $2\left(\frac{4\pi }{3}-\sqrt{3}\right)=\frac{8\pi }{3}-2\sqrt{3}$

Step-by-step solution

1. 1Intersections: $x^2=4\Rightarrow x=\pm 2$
2. 2Area = ${\int }_{-2}^2(4-x^2)dx={\left[4x-\frac{x^3}{3}\right]}_{-2}^2=\frac{16}{3}+\frac{16}{3}=\frac{32}{3}$