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If the area of the region \(x,y):1 + x^2leq yleq min \x + 7,11 - 3x\ \ is A, then 3A is equal to

Solution & Explanation

### Related Formula The area enclosed between upper bounding function y_textupper and lower function y_textlower is: textArea = int_a^b (y_textupper - y_textlower) \, dx ### Core Logic We need to find the intersection points of the curves to understand the min\x+7, 11-3x\ boundary transition: 1) x+7 = 11-3x implies 4x = 4 implies x = 1. Hence, the line switches behavior at x=1. 2) Intersecting 1+x^2 with x+7: x^2 - x - 6 = 0 implies (x-3)(x+2) = 0 implies x = -2 quad textor quad x = 3 3) Intersecting 1+x^2 with 11-3x: x^2 + 3x - 10 = 0 implies (x+5)(x-2) = 0 implies x = 2 quad textor quad x = -5
Area Under Curves diagram for Q57 - JEE Main 2025 Evening
Area Under Curves diagram for Q57 - JEE Main 2025 Evening
### Step 1: Set up Integrals The transition points show that from x = -2 to 1, the upper line is x+7, and from x = 1 to 2, the upper line is 11-3x. A = int_-2^1 ((x + 7) - (1 + x^2)) \, dx + int_1^2 ((11 - 3x) - (1 + x^2)) \, dx A = int_-2^1 (x + 6 - x^2) \, dx + int_1^2 (10 - 3x - x^2) \, dx ### Step 2: Integration Evaluation Evaluating the first integral: left[ fracx^22 + 6x - fracx^33 right]_-2^1 = left(frac12 + 6 - frac13right) - left(2 - 12 + frac83right) = frac376 - left(-frac223right) = frac272 Evaluating the second integral: left[ 10x - frac3x^22 - fracx^33 right]_1^2 = left(20 - 6 - frac83right) - left(10 - frac32 - frac13right) = frac343 - frac496 = frac196 Total Area A: A = frac272 + frac196 = frac81 + 196 = frac1006 = frac503 ### Step 3: Calculate 3A Multiplying the total area by 3: 3A = 3 cdot left(frac503right) = 50 ### Pattern Recognition When a boundary contains a min\\ or max\\ component, always solve for their internal intersection first to identify the exact splitting point of your definite integrals. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus

Reference Study Guides

Integral Calculus

More Integral Calculus Previous-Year Questions

Q57 2025 Properties of Definite Integrals
Let (a, b) be the point of intersection of the curve x^2 = 2y and the straight line y - 2x - 6 = 0 in the second quadrant. Then the integral I = int_a^b frac9x^21 + 5^x \, dx is equal to:
  • A. 24
  • B. 27
  • C. 18
  • D. 21

Solution

### Related Formula textKing's Property: int_a^b f(x) dx = int_a^b f(a+b-x) dx ### Core Logic First, we find the coordinates of intersection in the second quadrant to determine the integration limits a and b. ### Step 1: Find points of intersection Substitute y = fracx^22 into the line equation y - 2x - 6 = 0: fracx^22 - 2x - 6 = 0 implies x^2 - 4x - 12 = 0 (x - 6)(x + 2) = 0 implies x = 6 quad textor quad x = -2 Since the point (a, b) lies in the second quadrant, x must be negative: a = -2 b = 2(a) + 6 = 2(-2) + 6 = 2 Thus, the integration limits are a = -2 and b = 2. ### Step 2: Solve the Integral using King's Property The integral is: I = int_-2^2 frac9x^21 + 5^x dx quad text--- (1) Apply King's property, substituting x to -x (since -2 + 2 - x = -x): I = int_-2^2 frac9(-x)^21 + 5^-x dx = int_-2^2 frac9x^2 cdot 5^x1 + 5^x dx quad text--- (2) Adding equations (1) and (2): 2I = int_-2^2 9x^2 left( frac1 + 5^x1 + 5^x right) dx = int_-2^2 9x^2 dx Since 9x^2 is an even function: 2I = 2 int_0^2 9x^2 dx implies I = left[ 3x^3 right]_0^2 = 3(8) - 0 = 24 ### Pattern Recognition Whenever you see an exponential denominator like 1 + c^x inside a symmetric interval definite integral [-a, a], applying King's property will almost always cancel the exponential factor cleanly when the remaining numerator is even. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus
Q63 2025 Integration by Rationalisation
4int_0^1left(frac1sqrt3 + x^2 + sqrt1 + x^2right)mathrmdx - 3log_eleft(sqrt3right) is equal to:
  • A. 2 + sqrt2 + log_mathrmeleft(1 + sqrt2right)
  • B. 2 - sqrt2 - log_mathrmeleft(1 + sqrt2right)
  • C. 2 + sqrt2 - log_mathrmeleft(1 + sqrt2right)
  • D. 2 - sqrt2 + log_mathrmeleft(1 + sqrt2right)

Solution

### Related Formula int sqrta^2 + x^2 dx = fracx2 sqrta^2 + x^2 + fraca^22 lnleft| x + sqrta^2 + x^2 right| ### Core Logic We first rationalise the denominator to split the integral into two standard integration terms. ### Step 1: Rationalise the integrand Multiply the numerator and denominator by sqrt3+x^2 - sqrt1+x^2: frac1sqrt3 + x^2 + sqrt1 + x^2 = fracsqrt3 + x^2 - sqrt1 + x^2(3+x^2) - (1+x^2) = fracsqrt3 + x^2 - sqrt1 + x^22 Thus, the integral expression simplifies to: I = 4 int_0^1 left( fracsqrt3 + x^2 - sqrt1 + x^22 right) dx - 3log_eleft(sqrt3right) I = 2 int_0^1 sqrt3 + x^2 dx - 2 int_0^1 sqrt1 + x^2 dx - frac32 log_e 3 ### Step 2: Evaluate the integrals For the first integral: 2 int_0^1 sqrt3 + x^2 dx = 2 left[ fracx2 sqrt3 + x^2 + frac32 lnleft| x + sqrt3 + x^2 right| right]_0^1 = left[ x sqrt3 + x^2 + 3 lnleft| x + sqrt3 + x^2 right| right]_0^1 = left( sqrt4 + 3 ln(1 + sqrt4) right) - left( 0 + 3 lnsqrt3 right) = 2 + 3 ln 3 - frac32 ln 3 = 2 + frac32 ln 3 For the second integral: -2 int_0^1 sqrt1 + x^2 dx = -2 left[ fracx2 sqrt1 + x^2 + frac12 lnleft| x + sqrt1 + x^2 right| right]_0^1 = - left[ x sqrt1 + x^2 + lnleft| x + sqrt1 + x^2 right| right]_0^1 = - left( sqrt2 + ln(1 + sqrt2) right) ### Step 3: Sum the terms Now compile all terms: I = left( 2 + frac32 ln 3 right) - sqrt2 - ln(1 + sqrt2) - frac32 ln 3 I = 2 - sqrt2 - ln(1 + sqrt2) ### Pattern Recognition Integration of roots of quadratics: Always look for algebraic rationalisation when dealing with sum of root denominators. It directly reduces complex fractions into standard integrable functions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus
Q53 2025 Area Enclosed by Two Curves
Let the area enclosed between the curves |y| = 1 - x^2 and x^2 + y^2 = 1 be alpha. If 9alpha = beta pi + gamma, beta, gamma are integers, then the value of |beta - gamma| equals
  • A. 27
  • B. 18
  • C. 15
  • D. 33

Solution

### Related Formula Area enclosed between two symmetric functions across four quadrants: alpha = 4 times left( textArea of circle in 1^textst text quadrant - int_0^1 y_textparabola \, dx right) ### Core Logic The curves are symmetric across the axes. The region is enclosed between the unit circle x^2 + y^2 = 1 and the parabolas y = pm(1 - x^2).
Area Enclosed by Two Curves diagram for Q53 - JEE Main 2025 Evening
Area Enclosed by Two Curves diagram for Q53 - JEE Main 2025 Evening
By symmetry, the area alpha in the first quadrant is the area under the circle minus the area under the parabola from x=0 to x=1. ### Step 1: Calculate the Area Components Area of a circle quadrant with r=1 is fracpi4. Area under the parabola: int_0^1 (1 - x^2) \, dx = left[ x - fracx^33 right]_0^1 = 1 - frac13 = frac23 ### Step 2: Evaluate the Total Area alpha = 4 left[ fracpi4 - frac23 right] = pi - frac83 Multiplying both sides by 9: 9alpha = 9pi - 24 Comparing with 9alpha = betapi + gamma, we get: beta = 9, quad gamma = -24 Hence, |beta - gamma| = |9 - (-24)| = 33 ### Pattern Recognition Symmetry helps divide the integration work by 4. Recognizing standard shapes (like a circle sector) allows you to bypass complex integration entirely for half of the problem. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus
Q72 2025 Integration by Substitution
If int left(frac1x +frac1x^3right)left(sqrt[2]3x^-24 + x^-26right)dx = - frac alpha3 (alpha + 1) left(3 x ^ beta + x ^ gammaright) ^ frac alpha + 1alpha + C, mathbfx > 0, (alpha, beta, gamma in mathbfZ), where mathbfC is the constant of integration, then alpha + beta + gamma is equal to
Numerical Answer. Answer: 19 to 19

Solution

### Related Formula Standard power integration formula rule is: int u^n \, du = fracu^n+1n+1 + C ### Core Logic Rewrite the integral by adjusting powers inside the radical container: I = int left(frac1x^2 + frac1x^4right) left(frac3x + frac1x^3right)^frac123 \, dx Let t = frac3x + frac1x^3 implies dt = -3left(frac1x^2 + frac1x^4right)\,dx. ### Step 1: Integrate Substituting t into the equation: int fract^1/23 \, dt-3 = -frac13 cdot fract^24/2324/23 = -frac233(24) t^frac2423 Comparing parameters directly yields: alpha = 23, quad beta = -1, quad gamma = -3 alpha + beta + gamma = 23 - 1 - 3 = 19 ### Pattern Recognition Pull out high powers from the root factor to seamlessly reveal a matching f'(x) substitution pair on the outside. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus

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