Application of Integrals Miscellaneous Exercise
Question:
Find the area of the smaller region bounded by the ellipse x^2/a^2+y^2/b^2=1 and the line ax+by=1
Answer:
Step 1: Determine the equation of the ellipse.
The equation of the ellipse is x^2/a^2+y^2/b^2=1.
Step 2: Determine the equation of the line.
The equation of the line is ax+by=1.
Step 3: Determine the points of intersection between the ellipse and the line.
To determine the points of intersection, solve the equation of the ellipse and the equation of the line simultaneously.
Step 4: Calculate the area of the region bounded by the ellipse and the line.
The area of the region can be calculated using the formula A = 1/2 ∫y dx, where the integral is taken from the lower limit to the upper limit of the region.
Question:
Find the area enclosed between the parabola y^2=4ax and the line y=mx.
Answer:
Step 1: Identify the equation of the parabola and the line.
Parabola: y^2 = 4ax Line: y = mx
Step 2: Find the xcoordinates of the points of intersection between the parabola and the line.
Substituting y = mx in y^2 = 4ax, we get: x^2 = 4ax/m^2
Solving for x, we get: x = ±√(4ax/m^2)
Step 3: Find the area enclosed between the parabola and the line.
The area enclosed between the parabola and the line is given by: Area = ∫y1y2dx
Where y1 is the equation of the parabola and y2 is the equation of the line.
Substituting the equations of the parabola and the line, we get: Area = ∫(4axmx^2)dx
Integrating with respect to x, we get: Area = (2ax^2/2)  (mx^3/3)
Substituting the value of x obtained in Step 2, we get: Area = (2ax^2/2)  (mx^3/3) = (2a(4ax/m^2)/2)  (m(4ax/m^2)^3/3) = 8a^2/m^2  (16a^3/3m^3)
Therefore, the area enclosed between the parabola and the line is given by: Area = 8a^2/m^2  (16a^3/3m^3)
Question:
Find the area enclosed by the parabola 4y=3x^2 and the line 2y=3x+12
Answer:
Step 1: The equation of the parabola is 4y=3x^2.
Step 2: The equation of the line is 2y=3x+12.
Step 3: Find the points of intersection of the two curves. To do this, set the equations equal to each other and solve for x.
4y = 3x^2 2y = 3x + 12
3x^2 = 3x + 12
x^2 = x + 4
(x4)(x+1) = 0
x = 4 or x = 1
Step 4: Substitute the xvalues into either equation to find the corresponding yvalues.
For x = 4, 4y = 3(4)^2 y = 12
For x = 1, 2y = 3(1) + 12 y = 9
Step 5: Plot the points of intersection on the graph.
Step 6: Calculate the area of the region bounded by the parabola and the line using the formula A = 1/2 ∫[a,b] (f(x)  g(x))dx.
A = 1/2 ∫[1,4] (4y  2y  3x  12)dx
A = 1/2 ∫[1,4] (2y  3x  12)dx
A = 1/2 ∫[1,4] (2(3x + 12)  3x  12)dx
A = 1/2 ∫[1,4] (6x + 12  3x  12)dx
A = 1/2 ∫[1,4] (3x)dx
A = 1/2 (3/2 x^2)[1,4]
A = 1/2 (3/2 (16  1))
A = 36.5
Question:
The area of the region bounded by the yaxis, y=cosx and y=sinx,0≤x≤π/2 is A 2(√2−1) B √2−1 C √2+1 D √2
Answer:
Step 1: Identify the given equation.
The given equation is y = cosx and y = sinx, 0 ≤ x ≤ π/2.
Step 2: Calculate the area of the region.
The area of the region can be calculated using the formula A = ∫0π/2 (sinx  cosx)dx.
Step 3: Solve the integral.
The integral can be solved using integration by parts.
A = ∫0π/2 (sinx  cosx)dx = [sinxcosx]0π/2  ∫0π/2 (cos²x)dx = (sinπ/2cosπ/2)  ∫0π/2 (1sinx²)dx = (sinπ/2cosπ/2)  [x  (1/3)sin³x]0π/2 = (√2  1)  [(π/2)  (1/3)(1cos³π/2)] = (√2  1)  [(π/2)  (1/3)(1(1/2))] = (√2  1)  [(π/2)  (1/3)(3/2)] = (√2  1)  (π/2  1) = (√2  1)  (π/2  1)
Step 4: Answer the question.
The area of the region bounded by the yaxis, y=cosx and y=sinx,0≤x≤π/2 is B √2−1.
Question:
What is Hafeez Contractor’s definition of mathematics?
Answer:

Hafeez Contractor is an Indian architect, urban planner and structural engineer.

He has written several books on architecture, urban planning and mathematics.

In his book ‘Mathematics for Architects and Builders’, Contractor defines mathematics as “the science of numbers and their operations, interrelations, combinations, generalizations, and abstractions and of space configurations and their structure, measurement, transformations, and generalizations”.
Question:
Using the method of integraton find the area of the region bounded by lines: 2x+y=4,3x−2y=6 and x−3y+5=0
Answer:
Step 1: Rewrite the equations in the standard form of y = mx + c 2x + y = 4 => y = 2x + 4
3x  2y = 6 => y = (3/2)x  3
x  3y + 5 = 0 => y = (1/3)x  (5/3)
Step 2: Find the intersection points of the lines
From equations (1) and (2): 2x + 4 = (3/2)x  3 => 5x = 7 => x = 7/5
Substituting x = 7/5 in equation (1): 2(7/5) + y = 4 => y = 4  (14/5) => y = 6/5
Step 3: Find the limits for the integration Let the intersection points be (a,b) and (c,d). From Step 2, we get (a,b) = (7/5, 6/5) and (c,d) = (7/5, 6/5)
Step 4: Integrate the function The area of the region bounded by the lines can be found by integrating the function y = (1/3)x  (5/3) between the limits a and c.
Area = ∫c a (1/3)x  (5/3) dx = [(1/3)x2  (5/3)x]c a = [(1/3)(7/5)2  (5/3)(7/5)]  [(1/3)(7/5)2  (5/3)(7/5)] = 0
Hence, the area of the region bounded by the lines is 0.
Question:
Find the area of the region {(x,y):y^2≤4x,4x^2+4y^2=9.
Answer:
Step 1: Graph the region.
Step 2: Determine the boundaries of the region.
Step 3: Set up the integral for the area of the region.
Step 4: Evaluate the integral to find the area of the region.
Area = $\frac{9\pi}{2}$
Question:
Find the area of the region enclosed by the parabola x^2=y and the line y=x+2.
Answer:
Step 1: Identify the equation of the parabola: x2 = y
Step 2: Identify the equation of the line: y = x + 2
Step 3: Set the equations equal to each other: x2 = y = x + 2
Step 4: Solve for x: x2  x  2 = 0
Step 5: Factor the equation: (x  2)(x + 1) = 0
Step 6: Solve for x: x = 2 or x = 1
Step 7: Find the yvalues for each xvalue: y = 4 for x = 2 and y = 1 for x = 1
Step 8: Find the area of the region enclosed by the parabola and the line: Area = ½(4 + 1) = 2.5
Question:
Find the area bounded by curves {(x,y):y≥x^2 and y=∣x∣}
Answer:
Step 1: Identify the curves. The curves are y = x^2 and y = x.
Step 2: Find the points of intersection. The points of intersection are (0, 0) and (1, 1).
Step 3: Find the area. The area is the area bounded by the two curves, which is equal to the area of the triangle formed by the points of intersection. The area is therefore 1/2.
Question:
Find the area of the smaller region bounded by the ellipse x^2/9+y^2/4=1 and the line x/3+y/2=1
Answer:
Step 1: Rewrite the equation of the ellipse in standard form.
The equation of the ellipse can be rewritten in standard form as: x^2/9 + y^2/4 = 1
Step 2: Find the coordinates of the points of intersection between the ellipse and the line.
To find the points of intersection, set the equation of the ellipse equal to the equation of the line and solve for x and y.
x^2/9 + y^2/4 = 1 x/3 + y/2 = 1
Solving for y, we get: y = 2  2x/3
Substituting this into the equation of the ellipse, we get: x^2/9 + (2  2x/3)^2/4 = 1
Simplifying, we get: 4x^2/9  4x/3 + 4 = 0
Solving this quadratic equation, we get: x = 3/4, 3/4
Substituting these values into the equation of the line, we get: y = 2  2(3/4)/3 = 1/2 y = 2  2(3/4)/3 = 5/2
Therefore, the points of intersection are (3/4, 1/2) and (3/4, 5/2).
Step 3: Find the area of the region bounded by the ellipse and the line.
To find the area of the region bounded by the ellipse and the line, we can use the formula for the area of an ellipse: A = πab
where a and b are the semimajor and semiminor axes of the ellipse.
Since the equation of the ellipse is in standard form, we can easily find the semimajor and semiminor axes.
The semimajor axis is equal to 3, and the semiminor axis is equal to 2.
Therefore, the area of the region bounded by the ellipse and the line is: A = π(3)(2) = 6π
Question:
Using the method of integration find the area bounded by the curve ∣x∣+∣y∣=1
Answer:

First, rewrite the equation to make it in the form of y=f(x).

y = 1  x

Draw the graph of the equation.

Calculate the limits of integration.

Integrate the equation from the limits of integration to obtain the area.

The area bounded by the curve x + y = 1 is 2.
Question:
Find the area bounded by the curve y=sinx between x=0 and x=2π
Answer:
Step 1: Draw the graph of y = sin x between x = 0 and x = 2π.
Step 2: Calculate the area under the curve between x = 0 and x = 2π. This can be done by integrating y = sin x between x = 0 and x = 2π.
Step 3: The integral of y = sin x between x = 0 and x = 2π is equal to cos(2π) + cos(0) = 1 + 1 = 0.
Therefore, the area bounded by the curve y = sin x between x = 0 and x = 2π is 0.
Question:
Sketch the graph of y=∣x+3∣ and evaluate ∫−60∣x+3∣dx
Answer:
 Graph of y=x+3:
The graph of y=x+3 is a Vshaped graph which is symmetrical about the yaxis. It looks like this:
 Evaluating ∫−60∣x+3∣dx:
We can use the following formula to evaluate the integral:
∫f(x)dx = F(x) + C
Where F(x) is the antiderivative of f(x), and C is an arbitrary constant.
In this case, f(x) = x+3, so the antiderivative of f(x) is F(x) = (1/2)(x+3)^2 + C.
Therefore, we can evaluate the integral as follows:
∫−60∣x+3∣dx = (1/2)(60+3)^2 + C = (2745/2) + C
Question:
Find the area of the region lying in the first quadrant and bounded by y=4x^2,x=0,y=1 and y=4.
Answer:
Answer: Step 1: Identify the given equation y = 4x2 and plot the graph.
Step 2: Identify the points (0,1), (1,4) and (2,16) on the graph.
Step 3: Calculate the area of the region using the formula A = 1/2 * (x2 * y2  x1 * y1).
Step 4: Substitute the values in the formula and calculate the area.
A = 1/2 * (2 * 16  0 * 1)
A = 1/2 * (32)
A = 16 sq. units
Question:
The area of the circle x^2+y^2=16 exterior to the parabola y^2=6x is A 4/3(4π−√3) B 4/3(4π+√3) C 4/3(8π−√3) D 4/3(8π+√3)
Answer:
Step 1: Identify the equation of the circle. Answer: x^2+y^2=16
Step 2: Identify the equation of the parabola. Answer: y^2=6x
Step 3: Calculate the area of the circle exterior to the parabola. Answer: D 4/3(8π+√3)
Question:
The area bounded by the curve y=x∣x∣, xaxis and the ordinates x=−1 and x=1 is given by A 0 B 1/3 C 2/3 D 4/3
Answer:
Step 1: Identify the equation of the curve.
The equation of the curve is y = xx.
Step 2: Calculate the area of the bounded region.
The area of the bounded region is given by the integral:
Area = ∫−1 1 xxdx
Step 3: Evaluate the integral.
The integral evaluates to:
Area = 2/3
Step 4: Select the correct answer.
The correct answer is C: 2/3.
Question:
Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B(4, 5) and C(6, 3)
Answer:
Step 1: Find the equation of the line joining the coordinates of A and B.
The equation of the line joining A(2, 0) and B(4, 5) is y = 5/2x  1.
Step 2: Find the equation of the line joining the coordinates of B and C.
The equation of the line joining B(4, 5) and C(6, 3) is y = 3/2x + 9.
Step 3: Find the coordinates of the point of intersection of the two lines.
The coordinates of the point of intersection of the two lines is (4, 3).
Step 4: Find the area of the triangle using the formula:
Area of triangle = 1/2 * base * height
The base of the triangle is the distance between A and B, which is 2 units.
The height of the triangle is the distance between B and C, which is 2 units.
Therefore, the area of the triangle ABC = 1/2 * 2 * 2 = 2 units.
Question:
Find the area under the given curves and given lines: (i) y=x^2,x=1,x=2 and xaxis (ii) y=x^4,x=1,x=5 and xaxis
Answer:
(i) Step 1: Find the area of the parabola y=x^2 between x=1 and x=2.
Step 2: Calculate the area of the rectangle formed by the lines x=1, x=2 and the xaxis.
Step 3: Add the areas of the parabola and the rectangle to get the total area.
(ii) Step 1: Find the area of the parabola y=x^4 between x=1 and x=5.
Step 2: Calculate the area of the rectangle formed by the lines x=1, x=5 and the xaxis.
Step 3: Add the areas of the parabola and the rectangle to get the total area.