The area (in sq. units) of the region (x, y): 0 ≤ y ≤ 2|x| + 1, 0 ≤ y ≤ x² + 1, |x| ≤ 3$$\{(x, y): 0 \le y \le 2|x| + 1, 0 \le y \le x^2 + 1, |x| \le 3\}$$ is

Related Formula

The area between two curves y₁(x)$y_1(x)$ and y₂(x)$y_2(x)$ from x = a$x = a$ to x = b$x = b$ is given by:

Core Logic

Analyze the curves to locate intersection points :

• Curve 1: y = e^x$y = e^x$
• Curve 2: y = |e^x - 1|$y = |e^x - 1|$
• Boundary: y-axis (x = 0$x = 0$)

For x < 0$x < 0$, e^x < 1 ⇒ |e^x - 1| = 1 - e^x$e^x < 1 \Rightarrow |e^x - 1| = 1 - e^x$ .

Find intersection: e^x = 1 - e^x ⇒ 2e^x = 1 ⇒ e^x = (1)/(2) ⇒ x = -ln 2$e^x = 1 - e^x \Rightarrow 2e^x = 1 \Rightarrow e^x = \frac{1}{2} \Rightarrow x = -\ln 2$.

Step 1: Set up the Definite Integral

The integration spans from x = -ln 2$x = -\ln 2$ to x = 0$x = 0$. In this interval, e^x ≥ 1 - e^x$e^x \ge 1 - e^x$:

Step 2: Integration Evaluation

Integrate term-by-term :

Pattern Recognition

Modulus graphs split at their critical point (e^x = 1 ⇒ x = 0$e^x = 1 \Rightarrow x = 0$). Recognizing that the required segment lies strictly in the negative quadrant simplifies the definition of the upper and lower functions immediately.

Chapter Mix

Class 12 Mathematics: Application of Integrals

Related Formula

The net area between two intersecting functions y₂(x)$y_2(x)$ and y₁(x)$y_1(x)$ from lower limit a$a$ to upper limit b$b$ is evaluated using the definite integral:

Core Logic

Let's perform a horizontal shift transformation by defining X = x + 2$X = x + 2$ to simplify the expressions:

• Parabola: y = x² + 4x + 2 = (x+2)² - 2 y = X² - 2$y = x^2 + 4x + 2 = (x+2)^2 - 2 \implies y = X^2 - 2$
• Absolute curve: y = |x + 2| y = |X|$y = |x + 2| \implies y = |X|$

This coordinate translation preserves area completely while shifting the axis to the origin.

Step 1: Compute Points of Intersection

Find where the transformed curves intersect by equating the functions: |X|² - 2 = |X|$|X|^2 - 2 = |X|$

Since |X| ≥ 0$|X| \geq 0$, we discard |X| = -1$|X| = -1$. This gives |X| = 2 X = ± 2$|X| = 2 \implies X = \pm 2$.

Step 2: Set up and Evaluate the Transformed Integral

Since both curves are symmetric about the vertical line X = 0$X = 0$, we can compute the area for the positive half and double it:

Integrate the polynomial row-by-row:

Pattern Recognition

Applying a horizontal variable substitution matching X = x + ± c$X = x + ± c$ shifts complex quadratic forms to centered formats, which avoids tedious arithmetic across asymmetrical integration limits.

Chapter Mix

Class 12 Mathematics: Application of Integrals

Related Formula

Area integrating with respect to y$y$:

Standard integral:

Core Logic

The boundary curves are:

• x = (1)/(1+y²)$x = \frac{1}{1+y^2}$
• x = (y²)/(2)$x = \frac{y^2}{2}$

Find intersection points by setting x$x$ equal:

Since y$y$ is real, y² = 1 y = ± 1$y^2 = 1 \implies y = \pm 1$. When y = ± 1$y = \pm 1$, x = (1)/(2)$x = \frac{1}{2}$. Intersection points are ((1)/(2), 1)$\left(\frac{1}{2}, 1\right)$ and ((1)/(2), -1)$\left(\frac{1}{2}, -1\right)$.

Step 1: Set up and Compute Area Integral

Between y = -1$y = -1$ and y = 1$y = 1$, (1)/(1+y²) ≥ (y²)/(2)$\frac{1}{1+y^2} \ge \frac{y^2}{2}$.

Since the integrand is an even function of y$y$:

Pattern Recognition

Whenever curves are functions of y²$y^2$, integrating along the y$y$-axis avoids dealing with messy radical functions (square roots) and naturally accounts for symmetry across the x$x$-axis.

Chapter Mix

Class 12 Mathematics: Application of Integrals

Related Formula

Core Logic

Plotting the boundary lines and tracking intersection points yields a composite geometric region bounding a central area.

Step 1: Set up the integral pieces

The bounded region A$A$ can be conceptualized as a total bounding box/rectangle minus specific external integrals:

Step 2: Evaluate the Integral

Step 3: Calculate 6A

6A = 6 × (7)/(3) = 14$6A = 6 \times \frac{7}{3} = 14$

Pattern Recognition

When dealing with multiple absolute functions (|x|$|x|$, |x-1|$|x-1|$), check for coordinate mirror symmetry across vertical axes to slash total required calculus computations in half.

Chapter Mix

Class 12 Mathematics: Application of Integrals

Related Formula

Area enclosed between two intersecting curves y₁ = f(x)$y_1 = f(x)$ and y₂ = g(x)$y_2 = g(x)$ from boundary limits x = a$x = a$ to x = b$x = b$:

Core Logic

Given the two boundary curves:

• xy + 4y = 16 y(x+4) = 16 y = (16)/(x+4)$xy + 4y = 16 \implies y(x+4) = 16 \implies y = \frac{16}{x+4}$
• x + y = 6 y = 6 - x$x + y = 6 \implies y = 6 - x$

Find the intersection points by equating the two expressions for y$y$:

Step 1: Integral Formulation

Between x = -2$x = -2$ and x = 4$x = 4$, the line y = 6 - x$y = 6 - x$ lies above the curve y = (16)/(x+4)$y = \frac{16}{x+4}$. Therefore, the required enclosed area is:

Step 2: Apply Limits and Simplify

Substitute the upper limit x = 4$x = 4$:

Substitute the lower limit x = -2$x = -2$:

Subtract the lower limit value from the upper limit value:

Pattern Recognition

Sees: Area bounded by a straight line and a shifting rectangular hyperbola. Shortcut: The roots of the difference equation x² - 2x - 8 = 0$x^2 - 2x - 8 = 0$ directly give the limits. Always check graph orientations to place the upper linear equation before the curve equation inside the integral bracket to preserve absolute area values.

Chapter Mix

Class 12 Mathematics: Application of Integrals Class 11 Mathematics: Conic Sections (Hyperbola)