CBSE Class 12 Mathematics Application Of Integrals Worksheet Set 05

CBSE Class 12 Mathematics Application Of Integrals (4). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. Find the area bounded by the curves \(4y = 3x^2\) and \(2y = 3x + 12\).
Answer:

1) \(3x^2 = 4y\)
(.) parabola
(.) vertex \((0,0)\)
(.) open towards \(+\)ve \(y\)-axis

2) \(2y = 3x + 12\)
(.) line
(.) points \((0, 6)\) and \((-4, 0)\)

Intersection points:
Solving \(3x^2 = 4y\) and \(2y = 3x + 12\)
We have \((4, 12)\) and \((-2, 3)\)
Required area = \(\int_{-2}^{4} \left( \frac{3x+12}{2} \right) - \left( \frac{3x^2}{4} \right) \, dx\)
\[ = \frac{1}{4} \int_{-2}^{4} (6x + 24 - 3x^2) \, dx \]
\[ = \frac{3}{4} \int_{-2}^{4} (2x + 8 - x^2) \, dx \]
\[ = \frac{3}{4} \left[ x^2 + 8x - \frac{x^3}{3} \right]_{-2}^{4} \]
\[ = \frac{3}{4} \left[ \left(16 + 32 - \frac{64}{3}\right) - \left(4 - 16 + \frac{8}{3}\right) \right] \]
\[ = \frac{3}{4} \left[ \frac{80}{3} + \frac{28}{3} \right] = \frac{3}{4} \left( \frac{108}{3} \right) = \frac{108}{4} = 27 \]
\( \therefore \) Required area = \(27\) square units ans.

Question. Find the area bounded by \(y^2 = 4ax\) and \(y = mx\).
Answer:

1) \(y^2 = 4ax\)
(.) parabola
(.) vertex \((0,0)\)
(.) open towards \(+\)ve \(x\)-axis

2) \(y = mx\)
(.) line
(.) passing through \((0,0)\)

Intersection Points:
Solving \(y^2 = 4ax\) and \(y = mx\)
we have, \((0,0)\) and \(\left(\frac{4a}{m^2}, \frac{4a}{m}\right)\)
Required area = \(\int_{0}^{\frac{4a}{m^2}} (2\sqrt{a}\sqrt{x} - mx) \, dx\)
\[ = \left[ 2\sqrt{a} \cdot \frac{2}{3}x^{\frac{3}{2}} - \frac{mx^2}{2} \right]_{0}^{\frac{4a}{m^2}} \]
\[ = \left[ \frac{4\sqrt{a}}{3} \left(\frac{4a}{m^2}\right)^{\frac{3}{2}} - \frac{m}{2} \left(\frac{4a}{m^2}\right)^2 \right] - [0] \]
\[ = \frac{4\sqrt{a}}{3} \cdot \left(\frac{8a\sqrt{a}}{m^3}\right) - \frac{m}{2} \left(\frac{16a^2}{m^4}\right) \]
\[ = \frac{32a^2}{3m^3} - \frac{8a^2}{m^3} \]
\[ = \frac{8a^2}{3m^3} \]
\( \dots \) Required area \(\frac{8a^2}{3m^3}\) square units. ans.

Question. Find the area bounded by curve \(y^2 = 4a^2(x - 1)\) and lines \(x = 1\) and \(y = 4a\).
Answer:

1) \(x = 1\)
(.) line parallel to \(y\)-axis at \((1, 0)\)

2) \(y = 4a\)
(.) line parallel to \(x\)-axis at \((0, 4a)\)

3) \(y^2 = 4a^2(x - 1)\)
(.) shifting parabola
(.) vertex \((1,0)\)
(.) open towards \(+\)ve \(x\)-axis

Intersection Points:
Solving \(y^2 = 4a^2(x - 1)\) and \(y = 4a\)
We have \(16a^2 = 4a^2(x - 1)\)
\( 4 = x - 1 \Rightarrow x = 5 \therefore \) point \((5, 4a)\)
Required area = \(\int_{1}^{5} (4a - 2a\sqrt{x - 1}) \, dx\)
\[ = \left[ 4ax - 2a \cdot \frac{2}{3}(x - 1)^{\frac{3}{2}} \right]_{1}^{5} \]
\[ = \left[ 20a - \frac{4a}{3}(4)^{\frac{3}{2}} \right] - [4a - 0] \]
\[ = 20a - \frac{32a}{3} - 4a \]
\[ = 16a - \frac{32a}{3} \]
\[ = \frac{16a}{3} \]
\( \therefore \) Required area = \(\frac{16a}{3}\) sq. units. ans.

Question. Find the area of the region bounded by \(x^2 = 4y\), \(y = 2\), \(y = 4\) and \(y\)-axis.
Answer:

1) \(x^2 = 4y\)
(.) parabola
(.) vertex \((0,0)\)
(.) open towards \(+\)ve \(y\)-axis

2) \(y = 2\)
(.) line parallel to \(x\)-axis at \((0,2)\)

3) \(y = 4\)
(.) line parallel to \(x\)-axis at \((0,4)\)

4) \(y\)-axis

Intersection points:
Solving \(x^2 = 4y\) and \(y = 4\) we have \((4,4)\)
Solving \(x^2 = 4y\) and \(y = 2\) we have \((2\sqrt{2} , 2)\)
Required area = \(\int_{0}^{2\sqrt{2}} (4 - 2) \, dx + \int_{2\sqrt{2}}^{4} \left(4 - \frac{x^2}{4}\right) \, dx\)
\[ = (2x)_{0}^{2\sqrt{2}} + \left[ 4x - \frac{x^3}{12} \right]_{2\sqrt{2}}^{4} \]
\[ = 4\sqrt{2} + \left[ 16 - \frac{64}{12} \right] - \left[ 8\sqrt{2} - \frac{16\sqrt{2}}{12} \right] \]
\[ = 4\sqrt{2} + \frac{128}{12} - \frac{80\sqrt{2}}{12} \]
\[ = \frac{128-32\sqrt{2}}{12} \]
\[ = \frac{32-8\sqrt{2}}{3} \]
\( \therefore \) Required area = \(\frac{32-8\sqrt{2}}{3}\) sq. units

Question. Find the area of the region bounded by curves \(y = x^2 + 2\), \(y = x\), \(x = 0\) and \(x = 3\).
Answer:

1) \(y = x^2 + 2 \Rightarrow x^2 = y - 2\)
(.) shifting
(.) vertex \((0,2)\)
(.) open towards \(+\)ve \(y\)-axis

2) \(y = x\)
(.) line passing through \((0,0)\)

3) \(x = 0\)
(.) \(y\)-axis

4) \(x = 3\)
(.) line parallel to \(y\)-axis at \((3,0)\)

Intersection point:
Solving \(y = x^2 + 2\) and \(x = 3\)
We have \(x = 3\) & \(y = 11\)
\( \therefore (3,11) \)
Required area = \(\int_{0}^{3} (x^2 + 2 - x) \, dx\)
\[ = \left[ \frac{x^3}{3} + 2x - \frac{x^2}{2} \right]_{0}^{3} \]
\[ = \left(9 + 6 - \frac{9}{2}\right) - 0 \]
\[ = \frac{21}{2} \ ]
\( \therefore \) Required area = \(\frac{21}{2}\) square unit. ans.

Question. Find the area of the region \(0 \leq y \leq x^2 + 1\); \(0 \leq y \leq x + 1\); \(0 \leq x \leq 2\).
Answer:

1) \(y \geq 0\)

2) \(y \leq x^2 + 1 \Rightarrow x^2 \geq y - 1\)
(.) shifting parabola
(.) vertex \((0,1)\)
(.) open towards \(+\)ve \(y\)-axis
(.) solution outside the parabola

3) \(y \leq x + 1\)
(.) line passing through \((0,1)\) & \((-1,0)\)
(.) solution towards the origin

4) \(x \geq 0\)

5) \(x \leq 2\)
(.) line parallel to \(y\)-axis at \((2,0)\)
(.) solution towards the origin
\(x \geq 0\) & \(y \geq 0\) means solution in 1st quadrant.

Intersection points:
Solving \(y = x^2 + 1\) and \(y = x + 1\)
We have \((0,1)\) and \((1,2)\)
Required area = \(\int_{0}^{1} (x^2 + 1) \, dx + \int_{1}^{2} (x + 1) \, dx\)
\[ = \left[ \frac{x^3}{3} + x \right]_{0}^{1} + \left[ \frac{x^2}{2} + x \right]_{1}^{2} \]
\[ = \left[\frac{1}{3} + 1\right] + \left[(2 + 2) - \left(\frac{1}{2} + 1\right)\right] \]
\[ = \frac{4}{3} + 4 - \frac{3}{2} \]
\[ = \frac{8+24-9}{6} = \frac{23}{6} \]
\( \therefore \) Required area = \(\frac{23}{6}\) square units ans.

Question. Find the area bounded by the curves \(y = x|x|\), \(x\)-axis, \(x = -1\) and \(x = 1\).
Answer:

1) \(y = x|x| \Rightarrow\) two parabolas
(.) \(y = x^2\) ; \(x \geq 0\)
vertex \((0,0)\) open towards \(+\)ve \(y\)-axis
(.) \(y = -x^2\) ; \(x < 0\)
Vertex \((0,0)\) open towards \(-\)ve \(y\)-axis

2) \(x\)-axis

3) \(x = 1 \Rightarrow\) line parallel to \(y\)-axis at \((1,0)\)

4) \(x = -1 \Rightarrow\) line parallel to \(y\)-axis at \((-1,0)\)
Required area = \(\int_{-1}^{0} -(-x^2) \, dx + \int_{0}^{1} x^2 \, dx\)
\[ = \left[ \frac{x^3}{3} \right]_{-1}^{0} + \left[ \frac{x^3}{3} \right]_{0}^{1} \]
\[ = \left(0 - \left(-\frac{1}{3}\right)\right) + \left(\frac{1}{3} - 0\right) \]
\[ = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \]
\( \therefore \) Required area = \(\frac{2}{3}\) sq. units ans.

Question. The area between \(x = y^2\) and \(x = 4\) is divided in to two equal parts by line \(x = a\). Find value of \(a\).
Answer:

1) \(y^2 = x\)
(.) parabola, vertex \((0,0)\)
open towards \(+\)ve \(x\)-axis

2) \(x = 4\)
(.) line parallel to \(y\)-axis at \((4,0)\)

3) \(x = a\)
(.) line parallel to \(y\)-axis at \((a, 0)\)

Area of region A:
\[ = 2 \int_{0}^{a} (\sqrt{x} - 0) \, dx \quad \dots\{\text{due to symmetry}\} \]
\[ = 2 \cdot \frac{2}{3} \left[x^{\frac{3}{2}}\right]_{0}^{a} \]
\[ = \frac{4}{3} a\sqrt{a} \text{ sq. units} \]

Area of region B:
\[ = 2 \int_{a}^{4} (\sqrt{x} - 0) \, dx \quad \dots\{\text{due to symmetry}\} \]
\[ = 2 \cdot \frac{2}{3} \left[x^{\frac{3}{2}}\right]_{a}^{4} \]
\[ = \frac{4}{3} [8 - a^{\frac{3}{2}}] \text{ sq. units} \]

We are given that, area of region A = area of region B
\( \Rightarrow \frac{4}{3} a\sqrt{a} = \frac{4}{3} (8 - a^{\frac{3}{2}}) \)
\( \Rightarrow a^{\frac{3}{2}} = 8 - a^{\frac{3}{2}} \)
\( \Rightarrow 2a^{\frac{3}{2}} = 8 \)
\( \Rightarrow a^{\frac{3}{2}} = 4 \)
\( \Rightarrow a = 4^{\frac{2}{3}} \) ans.

Question. Prove that the curves \(y^2 = 4x\) and \(x^2 = 4y\) divide the area of the square bounded by lines \(x = 0\), \(y = 4\), \(y = 0\) and \(x = 4\) in to three equal parts.
Answer:

1) \(y^2 = 4x\)
(.) parabola, vertex \((0,0)\), open towards \(+\)ve \(x\)-axis

2) \(x^2 = 4y\)
(.) parabola, vertex \((0,0)\), open towards \(+\)ve \(y\)-axis

3) \(x = 0\)
(.) y-axis

4) \(x = 4\)
(.) line parallel to \(y\)-axis at \((4,0)\)

5) \(y = 0\)
(.) equation of \(x\)-axis

6) \(y = 4\)
(.) line parallel to \(x\)-axis at \((0,4)\)

Area of region A:
\[ = \int_{0}^{4} (4 - 2\sqrt{x}) \, dx \]
\[ = \left(4x - \frac{4}{3}x^{\frac{3}{2}}\right)_{0}^{4} \]
\[ = \left(16 - \frac{4}{3}(8)\right) - (0) \]
\[ = \frac{16}{3} \text{ sq. units} \]

Area of region B:
\[ = \int_{0}^{4} \left(2\sqrt{x} - \frac{x^2}{4}\right) \, dx \]
\[ = \left(\frac{4}{3}x^{\frac{3}{2}} - \frac{x^3}{12}\right)_{0}^{4} \]
\[ = \left(\frac{4}{3}(8) - \frac{64}{12}\right) - 0 \]
\[ = \frac{32}{3} - \frac{16}{3} = \frac{16}{3} \text{ sq. units} \]

Area of region C:
\[ = \int_{0}^{4} \left(\frac{x^2}{4} - 0\right) \, dx \]
\[ = \left(\frac{x^3}{12}\right)_{0}^{4} \]
\[ = \frac{64}{12} = \frac{16}{3} \text{ sq. units} \]
Clearly, the parabolas divide the area of the square in to three equal parts.

Question. Find the area of the region \(\{(x, y): x^2 \leq y \leq |x|\}\)
Answer:

1) \(x^2 \leq y\)
(.) parabola
(.) vertex \((0,0)\)
(.) open towards \(+\)ve \(y\)-axis
(.) solution inside the parabola

2) \(y \leq |x|\)
(.) \(y \leq x\); \(x \geq 0\) [line passes through \((0,0)\)]
(.) \(y \leq -x\); \(x < 0\) [line passes through \((0,0)\)]
Required area = \(2 \int_{0}^{1} (x - x^2) \, dx \quad \dots\{\text{due to symmetry}\}\)
\[ = 2 \left( \frac{x^2}{2} - \frac{x^3}{3} \right)_{0}^{1} \]
\[ = 2 \left( \frac{1}{2} - \frac{1}{3} \right) \]
\[ = 2 \left( \frac{1}{6} \right) = \frac{1}{3} \text{ sq. units ans.} \]