Application of Integrals: Miscellaneous Exercise Solutions & Concepts

Core Strategy for Solving Complex Area Problems

The Miscellaneous Exercise questions require a systematic approach. Unlike simpler exercises, you often need to combine multiple concepts. The first and most critical step is always to sketch the curves. A rough but accurate drawing helps you visualize the enclosed region, identify the upper and lower (or right and left) boundaries, and find the points of intersection. Once you have the sketch, you must decide whether to use vertical strips (integration with respect to x, i.e., ∫ y dx) or horizontal strips (integration with respect to y, i.e., ∫ x dy). Use vertical strips when you have clear 'top' and 'bottom' functions. Use horizontal strips when you have clear 'right' and 'left' functions. The limits of your integral are determined by the x or y coordinates of the points where the curves intersect. For complex regions, you might need to split the area into two or more parts and calculate each integral separately before adding them up.

Worked Examples from Miscellaneous Concepts

• Example 1: Find the area of the region bounded by the parabola y = x² and the line y = |x|. Step 1: Sketch the Curves. Draw the parabola y = x², which is a standard upward-opening parabola with its vertex at the origin. Draw the function y = |x|, which consists of two lines: y = x for x ≥ 0 and y = -x for x < 0. Notice the 'V' shape. Step 2: Identify the Bounded Region and Symmetry. The region is bounded in the first and second quadrants. The area in the first quadrant is identical to the area in the second quadrant due to the symmetry of both curves about the y-axis. We can find the area in the first quadrant and multiply it by 2. Step 3: Find Points of Intersection. In the first quadrant (x ≥ 0), y = |x| is y = x. Find the intersection of y = x² and y = x. x² = x => x² - x = 0 => x(x-1) = 0. So, x = 0 and x = 1. The intersection points are (0,0) and (1,1). Step 4: Set up the Integral. In the first quadrant, from x = 0 to x = 1, the line y = x is the upper curve and the parabola y = x² is the lower curve. Area in 1st quadrant = ∫[from 0 to 1] (Upper Curve - Lower Curve) dx = ∫[0 to 1] (x - x²) dx. Step 5: Evaluate the Integral. ∫(x - x²) dx = [x²/2 - x³/3]. Applying the limits [0 to 1]: (1²/2 - 1³/3) - (0²/2 - 0³/3) = 1/2 - 1/3 = 1/6. Step 6: Calculate Total Area. Total Area = 2 (Area in 1st quadrant) = 2 (1/6) = 1/3 sq. units.
• Example 2: Using integration, find the area of the triangular region whose vertices are A(-1, 0), B(1, 3), and C(3, 2). Step 1: Find Equations of the Lines (Sides of the Triangle). - Equation of AB: Using two-point form, (y - 0) = [(3-0)/(1-(-1))] (x - (-1)) => y = (3/2)(x+1) - Equation of BC: (y - 3) = [(2-3)/(3-1)] (x - 1) => y - 3 = (-1/2)(x-1) => y = (-1/2)x + 7/2 - Equation of AC: (y - 0) = [(2-0)/(3-(-1))] * (x - (-1)) => y = (2/4)(x+1) => y = (1/2)(x+1) Step 2: Sketch the Triangle and Plan the Integration. Plot the points A, B, C and draw the lines. The area can be found by splitting the region. Draw a vertical line from B(1,3) down to the x-axis. The total area is the area under line AB from x=-1 to x=1, plus the area under line BC from x=1 to x=3, minus the area under the line AC from x=-1 to x=3. Area = Area(under AB) + Area(under BC) - Area(under AC). Step 3: Set up the Integrals. Area = ∫[-1 to 1] (3/2)(x+1) dx + ∫[1 to 3] ((-1/2)x + 7/2) dx - ∫[-1 to 3] (1/2)(x+1) dx Step 4: Evaluate the Integrals. - ∫[-1 to 1] (3/2)(x+1) dx = (3/2)[x²/2 + x] from -1 to 1 = (3/2)[(1/2+1) - (1/2-1)] = (3/2)[3/2 - (-1/2)] = (3/2)[2] = 3 - ∫[1 to 3] ((-1/2)x + 7/2) dx = (1/2)[-x²/2 + 7x] from 1 to 3 = (1/2)[(-9/2+21) - (-1/2+7)] = (1/2)[33/2 - 13/2] = (1/2)[20/2] = 5 - ∫[-1 to 3] (1/2)(x+1) dx = (1/2)[x²/2 + x] from -1 to 3 = (1/2)[(9/2+3) - (1/2-1)] = (1/2)[15/2 - (-1/2)] = (1/2)[16/2] = 4 Step 5: Calculate the Final Area. Total Area = 3 + 5 - 4 = 4 sq. units.

Board Exam Strategy & Common Traps

1. A Neat Sketch is Non-Negotiable: Always start with a large, clear diagram. Label the axes, curves, and points of intersection. A correct diagram not only helps you set up the integral but also carries marks in the board exam.

2. Exploit Symmetry: Before jumping into calculations, check for symmetry. If the area is symmetric about the x-axis or y-axis, you can calculate the area in one quadrant and multiply it by 2 or 4. This significantly reduces calculation complexity and chances of error.

3. Master ∫(Upper - Lower) vs ∫(Right - Left): When using vertical strips (dx), the area is ∫ (y_upper - y_lower) dx. When using horizontal strips (dy), the area is ∫ (x_right - x_left) dy. Choosing the wrong setup is a common mistake. Pick the one that avoids splitting the integral, if possible.

4. Area is Always Positive: A definite integral can be negative if the area lies below the x-axis. However, the 'area' as a physical quantity must be positive. If you calculate an area and get a negative result, take its absolute value. For example, the area bounded by y = sin(x) from x=π to x=2π is ∫[π to 2π] sin(x) dx = [-cos(x)] from π to 2π = -cos(2π) - (-cos(π)) = -1 - (1) = -2. The area is |-2| = 2 sq. units.

Practice Questions with Solutions

• Q: Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x, and the circle x² + y² = 32. A: Step 1: Identify the curves and points of intersection. The curves are y=0 (x-axis), y=x, and x²+y²=32 (a circle with center (0,0) and radius √32 = 4√2). The line y=x intersects the circle when x² + x² = 32 => 2x² = 32 => x² = 16 => x=4 (since we are in the first quadrant). Step 2: Sketch the region. The region is bounded by the x-axis up to a point, then by the circle. The line y=x divides the area under the circle in the first quadrant. The point of intersection (4,4) is key. The line x=4 is the dividing line for our integration. Step 3: Set up the integrals. The area is split into two parts. From x=0 to x=4, the area is under the line y=x. From x=4 to x=4√2 (the circle's x-intercept), the area is under the circle y = √(32-x²). Area = ∫[0 to 4] x dx + ∫[4 to 4√2] √(32-x²) dx. Step 4: Evaluate the integrals. The first integral is [x²/2] from 0 to 4 = 16/2 = 8. The second integral uses the formula ∫√(a²-x²) dx = (x/2)√(a²-x²) + (a²/2)sin⁻¹(x/a). Here a²=32. ∫[4 to 4√2] √(32-x²) dx = [(x/2)√(32-x²) + 16sin⁻¹(x/4√2)] from 4 to 4√2. At x=4√2: (4√2/2)√(0) + 16sin⁻¹(1) = 16(π/2) = 8π. At x=4: (4/2)√(32-16) + 16sin⁻¹(4/4√2) = 2√16 + 16sin⁻¹(1/√2) = 2(4) + 16(π/4) = 8 + 4π. Value of second integral = 8π - (8 + 4π) = 4π - 8. Step 5: Sum the areas. Total Area = 8 + (4π - 8) = 4π. Final answer: 4π sq. units.
• Q: Find the area bounded by the ellipse x²/a² + y²/b² = 1 and the ordinates x = 0 and x = ae, where b² = a²(1-e²) and e < 1. A: Step 1: Isolate y from the ellipse equation. y²/b² = 1 - x²/a² => y² = (b²/a²)(a²-x²) => y = (b/a)√(a²-x²). We'll consider the area in the first quadrant and multiply by 2 for the area above the x-axis. Step 2: Set up the integral. We need to find the area from x=0 to x=ae in the first quadrant. Area (1st quad) = ∫[0 to ae] y dx = ∫[0 to ae] (b/a)√(a²-x²) dx. Step 3: Evaluate the integral. Take the constant (b/a) out. We use the standard formula ∫√(a²-x²) dx = (x/2)√(a²-x²) + (a²/2)sin⁻¹(x/a). Area (1st quad) = (b/a) [ (x/2)√(a²-x²) + (a²/2)sin⁻¹(x/a) ] from 0 to ae. Step 4: Apply the limits. At x=ae: (b/a) [ (ae/2)√(a²-a²e²) + (a²/2)sin⁻¹(ae/a) ] = (b/a) [ (ae/2)√(a²(1-e²)) + (a²/2)sin⁻¹(e) ]. Given b²=a²(1-e²), so √... = b. = (b/a) [ (ae/2)b + (a²/2)sin⁻¹(e) ] = ab(e/2) + (ab/2)sin⁻¹(e). At x=0, the expression is 0. Step 5: The question asks for the area bounded by the ellipse, which implies the total area in the upper and lower quadrants between these ordinates. So we multiply by 2. Total Area = 2 * [ ab(e/2) + (ab/2)sin⁻¹(e) ] = abe + ab sin⁻¹(e). Final answer: ab(e + sin⁻¹(e)) sq. units.
• Q: Find the area of the smaller region bounded by the ellipse x²/9 + y²/4 = 1 and the line x/3 + y/2 = 1. A: Step 1: Sketch the curves. The ellipse has semi-major axis a=3 and semi-minor axis b=2. The line x/3 + y/2 = 1 is the intercept form, crossing the x-axis at (3,0) and the y-axis at (0,2). These are the same points as the vertices of the ellipse in the first quadrant. Step 2: Identify the required area. The line forms a triangle with the axes in the first quadrant, and this triangle lies inside the ellipse. The 'smaller region' is the area of the ellipse in the first quadrant minus the area of the triangle formed by the line and the axes. Step 3: Plan the calculation using known formulas. This is faster than integration. Area of ellipse = πab. Area of triangle = (1/2) base height. Step 4: Calculate the area of the ellipse in the first quadrant. Total area of ellipse = π(3)(2) = 6π. Area in first quadrant = (1/4) 6π = 3π/2. Step 5: Calculate the area of the triangle. The triangle has base 3 and height 2. Area = (1/2) 3 * 2 = 3. Step 6: Find the required area. Required Area = (Area of ellipse in 1st quad) - (Area of triangle) = 3π/2 - 3. Final answer: (3π/2 - 3) sq. units. (This can also be solved by ∫(y_ellipse - y_line)dx from 0 to 3, but the geometric approach is much faster here).
• Q: Find the area enclosed between the parabola y² = 4ax and the line y = mx. A: Step 1: Find the points of intersection. Substitute y=mx into y²=4ax. (mx)² = 4ax => m²x² = 4ax => m²x² - 4ax = 0 => x(m²x - 4a) = 0. So, x=0 or x = 4a/m². The intersection points are (0,0) and (4a/m², 4a/m). Step 2: Sketch the curves. y²=4ax is a right-opening parabola. y=mx is a line through the origin. Assume m>0 for simplicity. The line is the upper curve if we integrate with respect to y, but it's easier to see the upper and lower curves with respect to x. Step 3: Set up the integral with respect to x. The upper curve is the parabola y = √4ax = 2√a√x. The lower curve is the line y=mx. The integration is from x=0 to x=4a/m². Wait, let's re-check the sketch. In the first quadrant, for small x, x > x², so y=mx is above y=√4ax only if m is very large. Let's check a point. At x=1, the parabola is at y=2√a and the line is at y=m. Let's assume the line is the upper curve. Area = ∫[0 to 4a/m²] (mx - 2√a√x) dx. This will give a negative result if our assumption is wrong, so we take the absolute value. Alternative (and better) view: The parabola is y = 2√a√x and the line is y=mx. The upper curve is the parabola and the lower curve is the line. Area = ∫[0 to 4a/m²] (2√a√x - mx) dx. Step 4: Evaluate the integral. Area = [2√a (x^(3/2))/(3/2) - m x²/2] from 0 to 4a/m². = [(4√a)/3 x^(3/2) - (m/2) x²] from 0 to 4a/m². = (4√a)/3 (4a/m²)^(3/2) - (m/2) (4a/m²)². = (4√a)/3 (8a^(3/2))/(m³) - (m/2) (16a²/m⁴). = (32a²)/(3m³) - (8a²)/m³ = (32a² - 24a²)/(3m³) = 8a²/(3m³). Final answer: 8a²/(3m³) sq. units.

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