JEE Advanced 1991

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: Sketch the curves and identify the region bounded by $x = \frac{1}{2}, x = 2, y = ln x$ and $y = 2^{x}$ . Find the area of this region.

Q: Wait, why is $2^x$ the upper curve? How do I know it doesn't intersect $\ln x$?

Great question! If you test a point in the interval, say $x=1$, $2^1 = 2$ and $\ln 1 = 0$. Since $2 > 0$, $2^x$ is above $\ln x$. Furthermore, $2^x$ is always positive and grows exponentially, while $\ln x$ is only defined for $x>0$ and grows very slowly. They do not intersect in this interval, so $2^x$ remains the upper boundary throughout.

Q: I always mess up the integration of $\ln x$. Is there a trick?

The trick is to treat $\ln x$ as $1 \cdot \ln x$ and use Integration by Parts. Let $u = \ln x$ and $dv = dx$. Then $du = \frac{1}{x} dx$ and $v = x$. The formula $\int u dv = uv - \int v du$ gives $x \ln x - \int x \cdot \frac{1}{x} dx$, which simplifies to $x \ln x - x$. Remembering this result is a huge time-saver for JEE!

Q: Why did the sign change when evaluating $\ln(\frac{1}{2})$?

Remember the logarithmic property $\ln(\frac{a}{b}) = \ln a - \ln b$. So, $\ln(\frac{1}{2}) = \ln 1 - \ln 2$. Since $\ln 1 = 0$, this becomes $-\ln 2$. When you substitute this into your expression, the negative sign from the formula and the negative sign from the logarithm property multiply to become positive. Always watch your signs during the subtraction step!

Visualized Solution

Visualizing the Curves

Identify the upper curve: $y = 2^{x}$ .
Identify the lower curve: $y = ln x$ .

Defining the Region

Vertical boundaries at $x = \frac{1}{2}$ and $x = 2$ .
The shaded region is the required area.

Area Formula Setup

Area $A = \int_{a}^{b} [f (x) - g (x)] d x$
Here, $f (x) = 2^{x}$ and $g (x) = ln x$ .

Raw Integral Setup

Area $A = \int_{\frac{1}{2}}^{2} (2^{x} - ln x) d x$

Splitting the Integral

$\int_{\frac{1}{2}}^{2} 2^{x} d x - \int_{\frac{1}{2}}^{2} ln x d x$

Integrating $2^{x}$

Standard formula: $\int a^{x} d x = \frac{a ^{x}}{l n a}$
So, $\int 2^{x} d x = \frac{2 ^{x}}{l n 2}$

Integrating $ln x$

Using Integration by Parts: $\int ln x \cdot 1 d x = x ln x - x$

Combined Antiderivative

Antiderivative: $F (x) = \frac{2 ^{x}}{l n 2} - (x ln x - x)$
Evaluate $[F (x)]_{\frac{1}{2}}^{2}$

Substituting Upper Limit ( $x = 2$ )

Substitute $x = 2$ :
$\frac{2 ^{2}}{l n 2} - (2 ln 2 - 2) = \frac{4}{l n 2} - 2 ln 2 + 2$

Substituting Lower Limit ( $x = \frac{1}{2}$ )

Substitute $x = \frac{1}{2}$ :
$\frac{2 ^{\frac{1}{2}}}{l n 2} - (\frac{1}{2} ln \frac{1}{2} - \frac{1}{2}) = \frac{2}{l n 2} - \frac{1}{2} ln \frac{1}{2} + \frac{1}{2}$

Simplifying $ln (\frac{1}{2})$

Property: $ln (\frac{1}{2}) = ln (2^{- 1}) = - ln 2$
Lower limit value becomes: $\frac{2}{l n 2} + \frac{1}{2} ln 2 + \frac{1}{2}$

Subtracting the Limits

Area $A = (\frac{4}{l n 2} - 2 ln 2 + 2) - (\frac{2}{l n 2} + \frac{1}{2} ln 2 + \frac{1}{2})$

Grouping Like Terms

Group terms: $(\frac{4}{l n 2} - \frac{2}{l n 2}) + (- 2 ln 2 - \frac{1}{2} ln 2) + (2 - \frac{1}{2})$

Final Answer

Final Area: $\frac{4 - 2}{l n 2} - \frac{5}{2} ln 2 + \frac{3}{2}$ sq. units.

00:00 / 00:00

The Sigma Insight: Area Bounded by Curves

Solution Diagram

Analyzing the Setup

To find the area

A

trapped between the curves

y = 2^{x}

and

y = ln x

within the interval

x \in [\frac{1}{2}, 2]

, we identify

2^{x}

as the upper function and

ln x

as the lower function.

The area is defined by the definite integral:

A = \int_{\frac{1}{2}}^{2} (2^{x} - ln x) d x

By the linearity of integration, we decompose this into two distinct integrals:

A = \int_{\frac{1}{2}}^{2} 2^{x} d x - \int_{\frac{1}{2}}^{2} ln x d x

Mastering the Antiderivatives

We evaluate the first integral using the standard form

\int a^{x} d x = \frac{a ^{x}}{l n a}

. This yields:

\int 2^{x} d x = \frac{2 ^{x}}{ln 2}

For the second integral, we apply integration by parts to find the antiderivative of

ln x

\int ln x d x = x ln x - x

Combining these results, the general antiderivative

F (x)

is:

F (x) = \frac{2 ^{x}}{ln 2} - (x ln x - x)

Final Calculation

We now evaluate

F (x)

at the boundaries

x = 2

and

x = \frac{1}{2}

. Substituting the upper limit

x = 2

F (2) = \frac{2 ^{2}}{ln 2} - (2 ln 2 - 2) = \frac{4}{ln 2} - 2 ln 2 + 2

Substituting the lower limit

x = \frac{1}{2}

and noting that

ln (\frac{1}{2}) = - ln 2

F (\frac{1}{2}) = \frac{2 ^{1/2}}{ln 2} - (\frac{1}{2} ln \frac{1}{2} - \frac{1}{2}) = \frac{2}{ln 2} + \frac{1}{2} ln 2 + \frac{1}{2}

Subtracting the lower limit from the upper limit, we obtain the final area:

A = (\frac{4}{ln 2} - 2 ln 2 + 2) - (\frac{2}{ln 2} + \frac{1}{2} ln 2 + \frac{1}{2})

Simplifying the expression, we arrive at the result:

A = \frac{4 - 2}{ln 2} - \frac{5}{2} ln 2 + \frac{3}{2}

Common Traps & FAQs

Wait, why is

2^{x}

the upper curve? How do I know it doesn't intersect

ln x

Great question! If you test a point in the interval, say

x = 1

2^{1} = 2

and

ln 1 = 0

. Since

2 > 0

2^{x}

is above

ln x

. Furthermore,

2^{x}

is always positive and grows exponentially, while

ln x

is only defined for

x > 0

and grows very slowly. They do not intersect in this interval, so

2^{x}

remains the upper boundary throughout.

I always mess up the integration of

ln x

. Is there a trick?

The trick is to treat

ln x

1 \cdot ln x

and use Integration by Parts. Let

u = ln x

and

d v = d x

. Then

d u = \frac{1}{x} d x

and

v = x

. The formula

\int u d v = uv - \int v d u

gives

x ln x - \int x \cdot \frac{1}{x} d x

, which simplifies to

x ln x - x

. Remembering this result is a huge time-saver for JEE!

Why did the sign change when evaluating

ln (\frac{1}{2})

Remember the logarithmic property

ln (\frac{a}{b}) = ln a - ln b

. So,

ln (\frac{1}{2}) = ln 1 - ln 2

. Since

ln 1 = 0

, this becomes

- ln 2

. When you substitute this into your expression, the negative sign from the formula and the negative sign from the logarithm property multiply to become positive. Always watch your signs during the subtraction step!

Animated Solution for Mathematics - Definite Integration: Sketch the curves and identify the region bounded by $x = \frac{1}{2}, x = 2, y = ln x$ and $y = 2^{x}$ . Find the area of this region.

Visualized Solution

Visualizing the Curves

Defining the Region

Area Formula Setup

Raw Integral Setup

Splitting the Integral

Integrating $2^{x}$

Integrating $ln x$

Combined Antiderivative

Substituting Upper Limit ( $x = 2$ )

Substituting Lower Limit ( $x = \frac{1}{2}$ )

Simplifying $ln (\frac{1}{2})$

Subtracting the Limits

Grouping Like Terms

Final Answer

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Mastering the Antiderivatives

Final Calculation

Similar Questions

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

Sketch the region bounded by the curves $y = x^{2}$ and $y = \frac{2}{1 + x ^{2}}$ . Find the area.

Find the area of the region bounded by the curves $y = x^{2}, y = 2 - x^{2}$ and $y = 2$ , which lies to the right of the line $x = 1$ .

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

Sketch the region bounded by the curves $y = 5 - x^{2}$ and $y = ∣ x - 1∣$ and find its area.

The area bounded by the curves $y = ln x, y = ln ∣ x ∣, y = ∣ ln x ∣$ and $y = ∣ ln ∣ x ∣∣$ is

Find the area bounded by the curves $x^{2} = y, x^{2} = - y$ and $y^{2} = 4 x - 3$ .

The area of the region bounded by the curves $y = ∣ x - 2∣, x = 1, x = 3$ and the x-axis is

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

Find the area bounded by the curve $x^{2} = 4 y$ and the straight line $x = 4 y - 2$ .

Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sketch the curves and identify the region bounded by x=21​,x=2,y=lnx and y=2x. Find the area of this region.

Visualized Solution

Visualizing the Curves

Defining the Region

Area Formula Setup

Raw Integral Setup

Splitting the Integral

Integrating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2x

Integrating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } lnx

Combined Antiderivative

Substituting Upper Limit ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2)

Substituting Lower Limit ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=21​)

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ln(21​)

Subtracting the Limits

Grouping Like Terms

Final Answer

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Mastering the Antiderivatives

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Compute the area of the region bounded by the curves y=exlnx and y=exlnx​ where lne=1.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sketch the region bounded by the curves y=x2 and y=1+x22​. Find the area.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the area of the region bounded by the curves y=x2,y=​2−x2​ and y=2, which lies to the right of the line x=1.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area enclosed by the curves y=loge​(x+e2), x=loge​(y2​) and x=loge​2, above the line y=1 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sketch the region bounded by the curves y=5−x2​ and y=∣x−1∣ and find its area.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area bounded by the curves y=lnx,y=ln∣x∣,y=∣lnx∣ and y=∣ln∣x∣∣ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the area bounded by the curves x2=y,x2=−y and y2=4x−3.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the region bounded by the curves y=∣x−2∣,x=1,x=3 and the x-axis is

.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 :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the area bounded by the curve x2=4y and the straight line x=4y−2.

Animated Solution for Mathematics - Definite Integration: Sketch the curves and identify the region bounded by $x = \frac{1}{2}, x = 2, y = ln x$ and $y = 2^{x}$ . Find the area of this region.

Integrating $2^{x}$

Integrating $ln x$

Substituting Upper Limit ( $x = 2$ )

Substituting Lower Limit ( $x = \frac{1}{2}$ )

Simplifying $ln (\frac{1}{2})$

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

Sketch the region bounded by the curves $y = x^{2}$ and $y = \frac{2}{1 + x ^{2}}$ . Find the area.

Find the area of the region bounded by the curves $y = x^{2}, y = 2 - x^{2}$ and $y = 2$ , which lies to the right of the line $x = 1$ .

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

Sketch the region bounded by the curves $y = 5 - x^{2}$ and $y = ∣ x - 1∣$ and find its area.

The area bounded by the curves $y = ln x, y = ln ∣ x ∣, y = ∣ ln x ∣$ and $y = ∣ ln ∣ x ∣∣$ is

Find the area bounded by the curves $x^{2} = y, x^{2} = - y$ and $y^{2} = 4 x - 3$ .

The area of the region bounded by the curves $y = ∣ x - 2∣, x = 1, x = 3$ and the x-axis is

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

Find the area bounded by the curve $x^{2} = 4 y$ and the straight line $x = 4 y - 2$ .