Trapezoidal Rule Area Calculation
To find the area under the curve $y = x^2 + 2x$ between $x = 0$ and $x = 4$ using the trapezoidal rule with a step size $h = 1$, we follow these steps:
1. Define Points and Calculate Function Values
The interval is $[0, 4]$ and the step size $h=1$. The points are $x_0 = 0, x_1 = 1, x_2 = 2, x_3 = 3, x_4 = 4$. We calculate the corresponding $y$ values using $y = x^2 + 2x$:
- $y_0 = f(0) = 0^2 + 2(0) = 0$
- $y_1 = f(1) = 1^2 + 2(1) = 1 + 2 = 3$
- $y_2 = f(2) = 2^2 + 2(2) = 4 + 4 = 8$
- $y_3 = f(3) = 3^2 + 2(3) = 9 + 6 = 15$
- $y_4 = f(4) = 4^2 + 2(4) = 16 + 8 = 24$
2. Apply Trapezoidal Rule Formula
The trapezoidal rule formula for approximating the area is:
Area $\approx \frac{h}{2} [y_0 + 2(y_1 + y_2 + ... + y_{n-1}) + y_n]$
Here, $h=1$ and $n=4$. Plugging in the values:
Area $\approx \frac{1}{2} [y_0 + 2(y_1 + y_2 + y_3) + y_4]$
Area $\approx \frac{1}{2} [0 + 2(3 + 8 + 15) + 24]$
3. Calculate the Area
Perform the calculation:
Area $\approx \frac{1}{2} [0 + 2(26) + 24]$
Area $\approx \frac{1}{2} [52 + 24]$
Area $\approx \frac{1}{2} [76]$
Area $\approx 38$
The calculated area using the trapezoidal rule is 38.
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
