Application of Integrals
1. Real Life Application - Why Area Matters? (Impactful Intro)
Imagine a dam holding water - how much water is stored? Area under inflow rate curve over time gives total water volume. Architecture: arch bridge shape is a curve, area under it tells material needed and load. Medicine: drug concentration in blood vs time, area under curve (AUC) tells total drug exposure to decide dose effectiveness. Economics: demand curve, area under it is consumer surplus total benefit. Daily life: car fuel consumption vs time area = total fuel used, speed vs time area = distance travelled, garden water flow rate vs time area = total water used. So integral ∫f(x)dx is not just maths, it is total accumulation. That is why application of integrals is powerful.
Real life: Dams water management ∫inflow dt=total water volume area=volume, Architecture arch profile y=curve area=material needed load distribution, Medicine drug concentration vs time AUC=total drug exposure dose effectiveness, Economics demand curve area=consumer surplus total benefit, Daily fuel consumption vs time total fuel, speed vs time distance=area, flow rate vs time total water. KEY IDEA ∫f(x)dx=area under curve=total accumulation.
• Dams & Water: ∫inflow dt=total water stored, Area=Total Water Volume
• Architecture: Arch y=curve area=material needed load
• Medicine: Drug concentration vs time AUC=total drug exposure dose monitoring
• Economics: Demand curve area=consumer surplus / total cost benefit
• Daily: Fuel vs time → total fuel, Speed vs time → distance=area, Flow rate vs time → total water
• KEY: ∫f(x)dx = area under curve = total accumulation
2. Area Under Simple Curve - Definite Integral as Area
Area under curve y=f(x) from x=a to b is ∫_a^b f(x)dx if f≥0. If curve above x-axis area positive. Example y=x² from 0 to1 area ∫0^1 x² dx = [x³/3]0^1=1/3. If curve below x-axis, integral negative but area is absolute value. If curve crosses, split into parts above and below. Geometrically area is limit of rectangles sum. This is base for all application. Remember definite integral is area with sign.
Area under y=f(x) a to b = ∫_a^b f(x)dx if f≥0. Example x² 0 to1 =1/3. If below x-axis area = |∫|. Crosses split. Definite integral = area with sign, absolute gives geometric area.
• Area y=f(x) from a to b = ∫_a^b f(x)dx if f≥0
• Example y=x² 0 to1 = [x³/3]0^1=1/3
• If below x-axis integral negative but geometric area positive absolute
• If crosses x-axis split into above/below parts
• Definite = area with sign, geometric = absolute
• Base for all area problems
3. Area Between Curve and Axes - Integrating w.r.t x or y
Sometimes curve is function of y, like x=g(y). Then area from y=c to d is ∫_c^d g(y)dy integrating w.r.t y. Example area bounded by y²=4ax and y-axis from y=0 to b? Or area under parabola from y-axis? Key idea: if strip is vertical dx, integrate w.r.t x: area = ∫ y dx. If strip horizontal dy, integrate w.r.t y: area = ∫ x dy. Choose easier orientation. For y²=4ax, y=√(4ax) area from 0 to a is ∫√(4ax)dx. Integration w.r.t y often easier for sideways parabola.
If vertical strip area = ∫ y dx from a to b. If horizontal strip area = ∫ x dy from c to d. For x=g(y) area = ∫_c^d g(y)dy. Choose orientation that makes strip simple. For y²=4ax use y=√(4ax) ∫ y dx.
• Vertical strip: Area = ∫ y dx from a to b (integrate w.r.t x)
• Horizontal strip: Area = ∫ x dy from c to d (w.r.t y)
• For x=g(y) area = ∫_c^d g(y)dy
• Choose orientation easier: sideways parabola easier w.r.t y
• y²=4ax → y=√(4ax) area 0 to a = ∫√(4ax)dx
• If strip touches y-axis use x dy
4. Area Bounded by Parabola y²=4ax and Line x=a - Standard Region
Classic problem: Area enclosed by parabola y²=4ax and line x=a (vertical line through focus). Parabola opens right, line x=a cuts it. Region symmetric about x-axis. Find intersection: put x=a in y²=4a·a=4a² y=±2a. Area = 2 × area above x-axis from 0 to a: =2∫_0^a √(4ax)dx =2·2√a ∫_0^a √x dx =4√a·[2/3 x^{3/2}]_0^a =8/3 a². Or using y integration: x from y²/4a to a. Area = ∫_-2a^{2a} (a - y²/4a)dy =8/3 a². Same answer. Remember 8/3 a² standard result.
Region y²=4ax and x=a symmetric about x-axis. Intersection y=±2a at x=a. Area =2∫_0^a √(4ax)dx=8/3 a². Or via y: ∫_-2a^{2a} (a - y²/4a)dy=8/3 a². Standard result 8/3 a².
• Region: Parabola y²=4ax + vertical line x=a
• Intersection: Put x=a → y²=4a² → y=±2a
• Symmetric about x-axis
• Area =2× area above axis =2∫_0^a √(4ax)dx=8/3 a²
• Via y: Area=∫_-2a^{2a} (a - y²/4a)dy=8/3 a²
• Standard result remember: 8/3 a²
5. Area of Circle and Ellipse Using Integration
Area of circle x²+y²=a² using integration: in first quadrant y=√(a²-x²) area quadrant =∫_0^a √(a²-x²)dx = πa²/4 using formula [x/2√(a²-x²)+a²/2 sin⁻¹(x/a)]. So total area =4×quadrant=πa². Similarly ellipse x²/a²+y²/b²=1 y=b/a√(a²-x²) quadrant area=πab/4 total=πab. These derivations prove known formulas via integration. Useful for half circle area problems too. Remember πab for ellipse, πa² for circle.
Circle x²+y²=a² y=√(a²-x²) quadrant ∫_0^a √(a²-x²)dx=πa²/4 total πa². Ellipse x²/a²+y²/b²=1 y=b/a√(a²-x²) quadrant πab/4 total πab. Formula ∫√(a²-x²)dx=x/2√...+a²/2 sin⁻¹(x/a).
• Circle x²+y²=a² y=√(a²-x²)
• Quadrant ∫_0^a √(a²-x²)dx = πa²/4 using √ formula
• Total circle =4×quadrant=πa²
• Ellipse x²/a²+y²/b²=1 y=b/a√(a²-x²)
• Quadrant =πab/4 total=πab
• Remember: Circle πa², Ellipse πab
• Use same formula for half circle
6. Area Between a Parabola and a Line - Intersection Method
General method: Area bounded by parabola y=x² and line y=x+2. Steps: 1) Find intersection points set equal x²=x+2 → x²-x-2=0 (x-2)(x+1)=0 x=2,-1 points (-1,1) and (2,4). 2) Identify top function line above parabola between -1 and 2. 3) Area=∫_-1^2 [(x+2)-x²]dx = [x²/2+2x-x³/3]_-1^2 =10/3 - (-7/6)=27/6=9/2=4.5. Key idea integrate upper minus lower across intersection limits. Draw rough graph to see which is upper. This method works for any two curves.
Steps: 1) Intersection set equal find limits, 2) Top minus bottom, 3) Area=∫_{x1}^{x2} [top-bottom]dx. Example y=x² y=x+2 intersect -1 and 2 points (-1,1)(2,4) Area=∫_-1^2 (x+2-x²)dx=9/2=4.5. Draw to know upper.
• Steps: 1) THE CURVES parabola y=x² line y=x+2
• 2) FIND INTERSECTION set equal x²=x+2 → x=2,-1 points (-1,1)(2,4)
• 3) SET UP Top-bottom Area=∫_-1^2 [(x+2)-x²]dx line above parabola
• 4) COMPUTE = [x²/2+2x-x³/3]_-1^2=10/3-(-7/6)=9/2
• AREA=9/2=4.5 square units
• KEY: Integrate (upper-lower) across intersection limits
7. Area Between Two Parabolas y²=4ax and x²=4ay
Two parabolas one opens right y²=4ax, other opens up x²=4ay. They intersect at (0,0) and (4a,4a) for same a? Actually if both a same, solve y²=4ax and x²=4ay → y=x²/4a put in first (x²/4a)²=4ax → x⁴=64a³x → x=0,4a y=0,4a. Area between them = area under y=√(4ax) minus area under y=x²/4a from 0 to4a =∫_0^{4a}[√(4ax)-x²/4a]dx = [2/3·...]. For a=1 result 16/3? Let's compute a=1: ∫0^4[2√x - x²/4]dx=[4/3 x^{3/2} - x³/12]0^4=4/3·8 -64/12=32/3-16/3=16/3. General 16a²/3. Symmetrical region.
Two parabolas y²=4ax (right) x²=4ay (up) intersect (0,0) and (4a,4a). Area between =∫_0^{4a}[√(4ax)-x²/4a]dx=16a²/3. For a=1 area 16/3. Draw both, upper is sqrt, lower is x²/4a.
• Curves: y²=4ax (right) and x²=4ay (up)
• Intersection: Solve → (0,0) and (4a,4a)
• Upper curve √ (4ax), lower curve x²/4a
• Area=∫_0^{4a}[√(4ax)-x²/4a]dx
• For a=1 Area=16/3 square units
• General 16a²/3 for same a
• Symmetrical lens shaped region
8. Area Between Two Curves - General Formula and Summary
General formula: Area between curves y=f(x) (upper) and y=g(x) (lower) from x=a to b where f≥g is ∫_a^b [f(x)-g(x)]dx. If curves cross, split at crossing points. If area bounded by x=g(y) and x=h(y) from y=c to d, area=∫_c^d [right-left]dy. Always draw rough graph, find intersection limits, decide top-bottom or right-left, integrate absolute difference. For summary: area under simple curve ∫y dx, between curve and y-axis ∫x dy, between two curves ∫(upper-lower)dx or ∫(right-left)dy, circle πa², ellipse πab, parabola x=a 8/3 a², two parabolas 16a²/3, parabola-line 9/2 for example.
General: Between y=f(x) top and y=g(x) bottom a to b Area=∫_a^b [f-g]dx. Between x=f(y) right and x=g(y) left c to d Area=∫_c^d [right-left]dy. If cross split. Always draw, find limits intersection, top-bottom right-left, integrate absolute. Summary formulas: simple ∫y dx, y-axis ∫x dy, two curves ∫(upper-lower), circle πa², ellipse πab, y²=4ax & x=a 8/3 a², two parabolas 16a²/3.
• General y: Top f(x) bottom g(x) Area=∫_a^b [f(x)-g(x)]dx
• General x: Right f(y) left g(y) Area=∫_c^d [right-left]dy
• If cross split at intersection
• Steps: Draw → Find intersection → Top/Bottom or Right/Left → Integrate absolute difference
• Circle x²+y²=a² Area πa², Ellipse πab
• Parabola y²=4ax & x=a Area 8/3 a²
• Two parabolas y²=4ax & x²=4ay Area 16a²/3
• Parabola y=x² & line y=x+2 Area 9/2 example
• Real Life: Dams water ∫inflow dt=total volume, Architecture arch area=material, Medicine drug AUC=total exposure, Economics demand area=consumer surplus, Daily fuel vs time total fuel, speed vs time distance=area
• Simple Curve: Area y=f(x) a to b = ∫_a^b f(x)dx if f≥0, absolute for below, split if cross
• w.r.t x vs y: Vertical strip Area=∫ y dx, Horizontal strip Area=∫ x dy, choose easier orientation
• Parabola y²=4ax & x=a: Intersection y=±2a Area 8/3 a² symmetric
• Circle & Ellipse: Circle x²+y²=a² Area πa², Ellipse x²/a²+y²/b²=1 Area πab via ∫√(a²-x²)dx formula
• Parabola & Line: Find intersection set equal, top-bottom, Area=∫(upper-lower)dx Example y=x² y=x+2 intersect -1,2 Area 9/2=4.5
• Two Parabolas: y²=4ax & x²=4ay intersect (0,0)(4a,4a) Area=∫_0^{4a}[√(4ax)-x²/4a]dx=16a²/3 (16/3 for a=1)
• General: Two curves top f bottom g Area=∫_a^b [f-g]dx, right-left ∫_c^d [right-left]dy, split if cross, draw rough graph first