The problem asks us to find the value of the function $f(1)$ given the area under the curve $y=f(x)$ from $x=0$ to $x=b$. We are given that the area, let's call it $A(b)$, is defined as:
$A(b) = \int_0^b f(x) dx = \sqrt{3+b^2} - \sqrt{3}$
Here, $f(x)$ is a continuous, non-negative function on $[0, 2]$, and $b \in (0, 2]$.
Relating Area and the Function using Calculus
According to the Fundamental Theorem of Calculus Part 1, the derivative of the definite integral function $A(b) = \int_0^b f(x) dx$ with respect to $b$ gives us the function $f(b)$.
So, we need to find the derivative of $A(b)$ with respect to $b$. $f(b) = \frac{d}{db} A(b) = \frac{d}{db} (\sqrt{3+b^2} - \sqrt{3})$
Calculating the Derivative
We differentiate the expression term by term:
- The derivative of the constant term $\sqrt{3}$ is 0.
- For the term $\sqrt{3+b^2}$, we use the chain rule. Let $u = 3+b^2$, then $\frac{du}{db} = 2b$. The derivative is $\frac{d}{db}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \frac{du}{db}$.
- Substituting back: $\frac{d}{db}(\sqrt{3+b^2}) = \frac{1}{2\sqrt{3+b^2}} \cdot (2b) = \frac{b}{\sqrt{3+b^2}}$.
Combining these, the derivative $f(b)$ is:
$f(b) = \frac{b}{\sqrt{3+b^2}} - 0 = \frac{b}{\sqrt{3+b^2}}$
Finding the Value of f(1)
Now we substitute $b=1$ into the expression for $f(b)$ to find $f(1)$.
$f(1) = \frac{1}{\sqrt{3+1^2}} = \frac{1}{\sqrt{3+1}} = \frac{1}{\sqrt{4}} = \frac{1}{2}$
The value of $f(1)$ is $\frac{1}{2}$, which is equal to 0.5.
Rounding to 1 decimal place, $f(1) = 0.5$.
Important Questions from Area Under Curve
- A function $y(x)$ is defined in the interval $[0, 1]$ on the x-axis as
$y(x) = \begin{cases} 2 & \text{if } 0 \le x < \frac{1}{3} \\ 3 & \text{if } \frac{1}{3} \le x < \frac{3}{4} \\ 1 & \text{if } \frac{3}{4} \le x \le 1 \end{cases}$
Which one of the following is the area under the curve for the interval $[0, 1]$ on the x-axis? - The area of the region bounded by the parabola $x = -y^2$ and the line $y = x + 2$ equals
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
In the figure shown above, PQRS is a square. The shaded portion is formed by the intersection of sectors of circles with radius equal to the side of the square and centers at S and Q.
The probability that any point picked randomly within the square falls in the shaded area is ___________.- If $f(x) = 2 \ln(\sqrt{e^x})$, what is the area bounded by $f(x)$ for the interval $[0, 2]$on the x-axis?
