Simplifying the Function $f(x)$
First, simplify the given function $f(x) = 2 \ln(\sqrt{e^x})$.
- Use the property $\sqrt{a} = a^{1/2}$: $f(x) = 2 \ln((e^x)^{1/2})$
- Use the exponent rule $(a^m)^n = a^{mn}$: $f(x) = 2 \ln(e^{x/2})$
- Use the logarithm property $\ln(e^a) = a$: $f(x) = 2 \times (x/2)$
- Simplify the expression: $f(x) = x$
The function simplifies to $f(x) = x$.
Calculating the Bounded Area
The area bounded by the function $f(x) = x$ on the interval $[0, 2]$ on the x-axis is calculated using a definite integral.
- Set up the integral for the area: Area = $\int_{0}^{2} f(x) dx$
- Substitute the simplified function $f(x) = x$: Area = $\int_{0}^{2} x dx$
- Find the antiderivative of $x$: The antiderivative is $\frac{x^2}{2}$.
- Evaluate the definite integral using the fundamental theorem of calculus: Area = $[\frac{x^2}{2}]_{0}^{2}$
- Substitute the upper and lower limits: Area = $\frac{2^2}{2} - \frac{0^2}{2}$
- Calculate the result: Area = $\frac{4}{2} - 0 = 2$
The area bounded by the function $f(x) = 2 \ln(\sqrt{e^x})$ over the interval $[0, 2]$ is 2.
Important Questions from Area Under Curve
- Let $I$ be the integral defined as follows: $$I = \int_{0}^{1} \int_{0}^{\sqrt{y}} dx dy + \int_{1}^{2} \int_{\sqrt{y-1}}^{1} dx dy$$ If the order of the integration is changed, then which one of the following is the correct expression for $I$?
Let $\alpha = \iint_S \vec{F} \cdot \hat{n} \, dS$, where $\vec{F} = (2x + 3z)\hat{i} + (xz - y)\hat{j} + (y^2 + 2z)\hat{k}$ and $S$ is the sphere with centre at $(3, -1, 2)$ and radius 9. Here, $\hat{n}$ is the unit normal drawn outward and $\hat{i}, \hat{j}, \hat{k}$ are unit vectors.
Then the value of $\frac{1}{36\pi} \alpha$ is equal to ________. (answer in integer)
The value of $\frac{4}{\pi} \int_0^{\pi/2} \sin^2 x \text{ dx}$ is _________________ (rounded off to two decimal places).
The work done by the force $F = (x + y)\hat{i} - (x^2 + y^2)\hat{j}$, where $\hat{i}$ and $\hat{j}$ are unit vectors in $\vec{OX}$ and $\vec{OY}$ directions, respectively, along the upper half of the circle $x^2 + y^2 = 1$ from $(1,0)$ to $(-1,0)$ in the $xy$-plane is
If the line $y = \alpha x$, $\alpha \geq \sqrt{2}$, divides the area of the region
$R: = \{(x, y) \in \mathbb{R}^2| 0 \leq x \leq \sqrt{y}, 0 \leq y \leq 2\}$
into two equal parts, then the value of $\alpha$ is equal to
