The area (in sq. units) of the region bounded by the curves y = 2^x and y = |x + 1|, in the first qu

Q: Why can we just remove the modulus sign? Is it always that simple?

In the first quadrant, we are given the constraint $x \ge 0$. Since $x + 1$ is always positive for any $x \ge 0$, the expression inside the modulus is already positive. Therefore, $|x + 1|$ is simply $x + 1$. It is a gift from the problem setter to simplify your path!

Q: How do I know there aren't other intersection points for $2^x = x + 1$?

That is a brilliant observation! $2^x$ is a strictly convex function, and $x + 1$ is a linear function. A convex function and a line can intersect at most at two points. We found them by inspection at $x = 0$ and $x = 1$. There are no others.

Q: Why is the line the upper curve? I'm always confused about this.

Always test a point! Pick a value in the interval $(0, 1)$, such as $x = 0.5$. The line gives $y = 0.5 + 1 = 1.5$. The curve gives $y = 2^{0.5} = \sqrt{2} \approx 1.414$. Since $1.5 > 1.414$, the line is clearly above the curve in this region.

Question

The area (in sq. units) of the region bounded by the curves $y = 2^x$ and $y = |x + 1|$, in the first quadrant is :

Options:

(A) 3/2 - 1/\log_e 2

(B) 1/2

(C) $\log_e 2 - 3/2$

(D) 3/2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the Cartesian plane, looking at the first quadrant. We are tasked with finding the area trapped between two mathematical entities: an exponential growth curve, $y = 2^x$, and a linear modulus function, $y = |x + 1|$.

This is not just a calculation; it is a story of how constraints simplify our world.

### Taming the Modulus

The modulus function often strikes fear into the hearts of students, but here, the 'First Quadrant' constraint is our savior. Because we are restricted to $x \ge 0$, the term $(x + 1)$ is always positive.

Thus, the modulus sign vanishes like magic, leaving us with the elegant linear equation $y = x + 1$. We have transformed a complex absolute value problem into a simple geometry problem.

### The Intersection Points

Now, we must find where these two paths cross. We set the equations equal to each other:

$$2^x = x + 1$$

This is a transcendental equation—one that does not yield to standard algebraic manipulation. However, in the JEE, intuition is a powerful tool.

Let us test the simplest integers. At $x = 0$, both sides equal $1$. At $x = 1$, both sides equal $2$.

We have found our boundaries: $x = 0$ and $x = 1$. These are the gates to our region.

### The Integral Setup

To find the area, we need to know which curve sits on top. By testing $x = 0.5$, we see that the line $y = x + 1$ yields $1.5$, while the curve $y = 2^x$ yields $\sqrt{2} \approx 1.414$.

The line is the ceiling, and the curve is the floor. The area $A$ is the integral of the difference:

$$A = \int_{0}^{1} ((x + 1) - 2^x) dx$$

### The Final Integration

Now, we integrate term by term. The integral of $x$ is $\frac{x^2}{2}$, the integral of $1$ is $x$, and the integral of $2^x$ is $\frac{2^x}{\ln 2}$.

Our antiderivative is:

$$\left[ \frac{x^2}{2} + x - \frac{2^x}{\ln 2} ight]_0^1$$

Substituting the upper limit $x = 1$:

$$\left( \frac{1}{2} + 1 - \frac{2}{\ln 2} ight) = \frac{3}{2} - \frac{2}{\ln 2}$$

Substituting the lower limit $x = 0$:

$$\left( 0 + 0 - \frac{1}{\ln 2} ight) = -\frac{1}{\ln 2}$$

Subtracting the lower limit from the upper limit, the negatives cancel out, leaving us with the final, elegant result:

**$$A = \frac{3}{2} - \frac{1}{\ln 2} 	ext{ square units}$$**

You have successfully navigated the curves and conquered the integral. Keep this confidence for your next challenge!

Animated Solution for Mathematics - Definite Integration: The area (in sq. units) of the region bounded by the curves $y = 2^{x}$ and $y = ∣ x + 1∣$ , in the first quadrant is :

Select Answer:

Visualized Solution

Analyzing the Given Curves

Simplifying the Modulus Function

Plotting the Exponential Curve

Equating the Curves

First Intersection Point

Second Intersection Point

The Bounded Region

Setting up the Area Integral

Performing the Integration

Substituting the Upper Limit

Substituting the Lower Limit

Calculating the Final Area

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Taming the Modulus

The Intersection Points

The Integral Setup

The Final Integration

Similar Questions

The area bounded by the curves $y = ∣ x ∣ - 1$ and $y = - ∣ x ∣ + 1$ is

The area of the region enclosed by the curves $y = e^{x}$ , $y = ∣ e^{x} - 1∣$ and y-axis is:

The area bounded by the curves $y = ∣ x^{2} - 1∣$ and $y = 1$ is

The area enclosed by the curves $y = lo g_{e} (x + e^{2})$ , $x = lo g_{e} (\frac{2}{y})$ and $x = lo g_{e} 2$ , above the line $y = 1$ is

The area enclosed between the curves $y^{2} = x$ and $y = ∣ x ∣$ is

The area of the region bounded by the curves $y = ∣ x - 1∣$ and $y = 3 - ∣ x ∣$ is

The area of the region bounded by the curves $x (1 + y^{2}) = 1$ and $y^{2} = 2 x$ is:

Compute the area of the region bounded by the curves $y = e x ln x$ and $y = \frac{l n x}{e x}$ where $ln e = 1$ .

Let the functions $f : R \to R$ and $g : R \to R$ be defined by $f (x) = e^{x - 1} - e^{- ∣ x - 1∣}$ and $g (x) = \frac{1}{2} (e^{x - 1} + e^{1 - x})$ . Then the area of the region in the first quadrant bounded by the curves $y = f (x)$ , $y = g (x)$ and $x = 0$ is

The area enclosed between the curves $y = x ∣ x ∣$ and $y = x - ∣ x ∣$ is :

Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area (in sq. units) of the region bounded by the curves y=2x and y=∣x+1∣, in the first quadrant is :