If the area of the region (x,y):1 + x²≤ y≤ x + 7,11 - 3x$\{(x,y):1 + x^2\leq y\leq \min \{x + 7,11 - 3x\} \}$ is A, then 3A is equal to

Related Formula

Core Logic

Let I = ∫-π/6π/6 π+4x¹¹1- (|x|+π/6)dx$I = \int_{-\pi/6}^{\pi/6}\frac{\pi+4x^{11}}{1-\sin(|x|+\pi/6)}dx$. The denominator 1 - (|x| + π/6)$1 - \sin(|x| + \pi/6)$ is an even function. The numerator can be split into an even part (π$\pi$) and an odd part (4x¹¹$4x^{11}$).

Step 1: Simplify to Even Integral

We are left with the even part:

Using even function property ∫₋ₐa f(x) dx = 2 ∫₀a f(x) dx$\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$:

Step 2: Substitution

Let t = x + (π)/(6) ⇒ dt = dx$t = x + \frac{\pi}{6} \Rightarrow dt = dx$. Limits: when x = 0 ⇒ t = π/6$x = 0 \Rightarrow t = \pi/6$, when x = π/6 ⇒ t = π/3$x = \pi/6 \Rightarrow t = \pi/3$.

Step 3: Solve the Integral

Rationalize the denominator:

Integrate directly:

Evaluate limits: Upper limit (π/3$\pi/3$): (π/3) + (π/3) = √(3) + 2$\tan(\pi/3) + \sec(\pi/3) = \sqrt{3} + 2$ Lower limit (π/6$\pi/6$): (π/6) + (π/6) = 1√(3) + 2√(3) = 3√(3) = √(3)$\tan(\pi/6) + \sec(\pi/6) = \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$

Pattern Recognition

Symmetric limits [-a, a]$[-a, a]$ instantly demand testing for odd/even parity. Any mixed polynomial like c + k xodd$c + k x^{\text{odd}}$ over an even denominator guarantees the odd power term strictly vanishes, halving calculation time.

Chapter Mix

Class 12 Maths: Definite Integration Class 11 Maths: Trigonometric Functions

Related Formula

Core Logic

Let I = 6∫₀π| 3x + x + 2x| dx$I = 6\int_{0}^{\pi}|\sin 3x + \sin x + \sin 2x| dx$. Apply sum-to-product on 3x + x$\sin 3x + \sin x$: 3x + x = 2 (2x) (x)$\sin 3x + \sin x = 2 \sin(2x) \cos(x)$

So the expression becomes: |2 (2x) x + 2x| = | 2x (2 x + 1)| = |2 x x (2 x + 1)|$|2 \sin(2x) \cos x + \sin 2x| = |\sin 2x (2 \cos x + 1)| = |2 \sin x \cos x (2 \cos x + 1)|$ Since x in [0, π]$x \in [0, \pi]$, x ≥ 0$\sin x \geq 0$. We can pull it out of the modulus. I = 12 ∫₀π x |2 ² x + x| dx$I = 12 \int_{0}^{\pi} \sin x |2 \cos^2 x + \cos x| dx$

Step 1: Coordinate Substitution

Substitute t = x$t = \cos x$, then dt = - x dx$dt = -\sin x dx$. Limits: when x = 0$x = 0$, t = 1$t = 1$. When x = π$x = \pi$, t = -1$t = -1$.

Step 2: Resolve Modulus Intervals

The roots of 2t² + t = 0$2t^2 + t = 0$ are t = 0$t = 0$ and t = -1/2$t = -1/2$. The quadratic 2t² + t$2t^2 + t$ is negative in the interval (-1/2, 0)$(-1/2, 0)$ and positive elsewhere. Split the integral:

Step 3: Evaluate Integrals

Anti-derivative: F(t) = (2t³)/(3) + (t²)/(2)$F(t) = \frac{2t^3}{3} + \frac{t^2}{2}$. F(1) = 2/3 + 1/2 = 7/6$F(1) = 2/3 + 1/2 = 7/6$ F(0) = 0$F(0) = 0$ F(-1/2) = 2(-1/8)/3 + 1/8 = -1/12 + 1/8 = 1/24$F(-1/2) = 2(-1/8)/3 + 1/8 = -1/12 + 1/8 = 1/24$ F(-1) = -2/3 + 1/2 = -1/6$F(-1) = -2/3 + 1/2 = -1/6$

Evaluate each segment:

• ∫₋₁-1/2 = F(-1/2) - F(-1) = 1/24 - (-1/6) = 1/24 + 4/24 = 5/24$\int_{-1}^{-1/2} = F(-1/2) - F(-1) = 1/24 - (-1/6) = 1/24 + 4/24 = 5/24$
• -∫-1/2⁰ = -(F(0) - F(-1/2)) = -(0 - 1/24) = 1/24$-\int_{-1/2}^{0} = -(F(0) - F(-1/2)) = -(0 - 1/24) = 1/24$
• ∫₀¹ = F(1) - F(0) = 7/6 - 0 = 28/24$\int_{0}^{1} = F(1) - F(0) = 7/6 - 0 = 28/24$

Sum of parts inside bracket: (5)/(24) + (1)/(24) + (28)/(24) = (34)/(24) = (17)/(12)$\frac{5}{24} + \frac{1}{24} + \frac{28}{24} = \frac{34}{24} = \frac{17}{12}$

Step 4: Final Output

Pattern Recognition

Whenever an integral features a cascading sum of sine frequencies like (kx)$\sin(kx)$, pair the highest and lowest frequencies first. The resulting common factor often matches the middle term, instantly yielding a clean polynomial substitution under t = x$t = \cos x$.

Chapter Mix

Class 12 Maths: Definite Integration Class 11 Maths: Trigonometric Functions

Related Formula

Core Logic

Let I = ∫₀¹ ⁻¹(1-2x+4x²) dx$I = \int_{0}^{1} \cot^{-1}(1-2x+4x^2) dx$. Convert ⁻¹$\cot^{-1}$ to ⁻¹$\tan^{-1}$:

Notice that 2x - (2x-1) = 1$2x - (2x-1) = 1$. Thus, the integrand is ⁻¹( (2x - (2x-1))/(1 + 2x(2x-1)) )$\tan^{-1}\left( \frac{2x - (2x-1)}{1 + 2x(2x-1)} \right)$.

Step 1: Apply Definite Integral Properties

Applying King's property to the second term ∫₀¹ ⁻¹(2x-1) dx$\int_0^1 \tan^{-1}(2x-1) dx$:

Wait, this implies ∫₀¹ ⁻¹(2x-1) dx = 0$\int_0^1 \tan^{-1}(2x-1) dx = 0$! Thus, I = ∫₀¹ ⁻¹(2x) dx$I = \int_0^1 \tan^{-1}(2x) dx$.

Step 2: Integration by Parts

Solve ∫₀¹ ⁻¹(2x) · 1 dx$\int_0^1 \tan^{-1}(2x) \cdot 1 \, dx$:

Let 1+4x² = t 8x dx = dt$1+4x^2 = t \implies 8x dx = dt$. At x=0, t=1$x=0, t=1$; at x=1, t=5$x=1, t=5$.

Step 3: Compare and Calculate Result

The original integral has a factor of 4$4$:

Compare with a ⁻¹(2) - b ₑ(5)$a\tan^{-1}(2) - b\log_e(5)$:

Therefore, 2a + b = 2(4) + 1 = 9$2a + b = 2(4) + 1 = 9$.

Pattern Recognition

Always convert ⁻¹$\cot^{-1}$ quadratic inputs into ⁻¹(x-y)/(1+xy)$\tan^{-1}\frac{x-y}{1+xy}$ forms. Apply King's property on symmetric limits; often one piece vanishes entirely.

Chapter Mix

Class 12 Maths: Definite Integrals Class 12 Maths: Inverse Trigonometric Functions

Related Formula

Core Logic

The integral bounds are from -π/2 ≈ -1.57$-\pi/2 \approx -1.57$ to π/2 ≈ 1.57$\pi/2 \approx 1.57$. We must split the integral at every integer point between these bounds.

Intervals:

• [-π/2, -1) [x] = -2$[-\pi/2, -1) \implies [x] = -2$
• [-1, 0) [x] = -1$[-1, 0) \implies [x] = -1$
• [0, 1) [x] = 0$[0, 1) \implies [x] = 0$
• [1, π/2) [x] = 1$[1, \pi/2) \implies [x] = 1$

Step 1: Splitting the Integral

Step 2: Evaluating Sub-integrals

Evaluate the limits for each constant integral:

Step 3: Simplifying the Expression

Group the ( (π)/(2) - 1 )$\left( \frac{\pi}{2} - 1 \right)$ terms:

Find a common denominator for the constants (LCD is 60):

Rewrite (7π)/(20)$\frac{7\pi}{20}$ with denominator 60:

So,

Pattern Recognition

Integration of step functions always transforms into a simple sum of rectangle areas (cᵢ × Δ xᵢ)$(c_i \times \Delta x_i)$. Immediately break the bounds at integers, turning a calculus problem into elementary arithmetic.

Chapter Mix

Class 12 Maths: Definite Integration

Related Formula

Core Logic

Integral: I = ∫ -11/2 x -5/2 x dx$I = \int \sin^{-11/2} x \cos^{-5/2} x \, dx$

The sum of powers is -(11)/(2) - (5)/(2) = -8$-\frac{11}{2} - \frac{5}{2} = -8$. Convert the integrand entirely into terms of x$\tan x$ and x$\sec x$ by dividing and multiplying by -11/2 x$\cos^{-11/2} x$.

Step 1: Integration by Substitution

Rewrite ⁸ x = ( ² x)³ ² x = (1 + ² x)³ ² x$\sec^8 x = (\sec^2 x)^3 \sec^2 x = (1 + \tan^2 x)^3 \sec^2 x$.

Substitute t = x dt = ² x dx$t = \tan x \implies dt = \sec^2 x \, dx$.

Expand (1 + t²)³ = 1 + 3t² + 3t⁴ + t⁶$(1 + t^2)^3 = 1 + 3t^2 + 3t^4 + t^6$.

Step 2: Evaluating the Anti-derivatives

Integrate term by term:

Substitute back t = x = (1)/( x)$t = \tan x = \frac{1}{\cot x}$, which implies t-a = ( x)^a$t^{-a} = (\cot x)^a$.

Step 3: Variable Assignment

Comparing with -(p₁)/(q₁)( x)9/2 - (p₂)/(q₂)( x)5/2 - (p₃)/(q₃)( x)1/2 + (p₄)/(q₄)( x)-3/2$-\frac{p_1}{q_1}(\cot x)^{9/2} - \frac{p_2}{q_2}(\cot x)^{5/2} - \frac{p_3}{q_3}(\cot x)^{1/2} + \frac{p_4}{q_4}(\cot x)^{-3/2}$:

p₁ = 2, q₁ = 9$p_1 = 2, q_1 = 9$ p₂ = 6, q₂ = 5$p_2 = 6, q_2 = 5$ p₃ = 6, q₃ = 1$p_3 = 6, q_3 = 1$ p₄ = 2, q₄ = 3$p_4 = 2, q_4 = 3$

Calculate (15 p₁ p₂ p₃ p₄)/(q₁ q₂ q₃ q₄)$\frac{15 p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4}$:

Pattern Recognition

If ∫ ^m x ⁿ x dx$\int \sin^m x \cos^n x \,dx$ has m+n$m+n$ as a negative even integer, unconditionally extract |m+n|x$\sec^{|m+n|}x$ and set x = t$\tan x = t$. The expansion expands gracefully into polynomial power rules without any trig substitution hassle.

Chapter Mix

Class 12 Maths: Indefinite Integration