The value of $\frac{4}{\pi} \int_0^{\pi/2} \sin^2 x \text{ dx}$ is _________________ (rounded off to two decimal places).
Integral Evaluation: $\frac{4}{\pi} \int_0^{\pi/2} \sin^2 x \text{ dx}$
To find the value of the given expression, we need to evaluate the definite integral $\int_0^{\pi/2} \sin^2 x \text{ dx}$ and then multiply it by $\frac{4}{\pi}$.
Integral Step-by-Step
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Apply Trigonometric Identity: Use the identity $\sin^2 x = \frac{1 - \cos(2x)}{2}$.
The integral becomes:
$ \int_0^{\pi/2} \frac{1 - \cos(2x)}{2} \text{ dx} $
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Integrate the Expression: Find the antiderivative of the function.
$ \frac{1}{2} \int_0^{\pi/2} (1 - \cos(2x)) \text{ dx} = \frac{1}{2} \left[ x - \frac{\sin(2x)}{2} \right]_0^{\pi/2} $
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Apply Limits of Integration: Substitute the upper and lower limits.
$ \frac{1}{2} \left[ \left( \frac{\pi}{2} - \frac{\sin(\pi)}{2} \right) - \left( 0 - \frac{\sin(0)}{2} \right) \right] $
Since $\sin(\pi) = 0$ and $\sin(0) = 0$, this simplifies to:$ \frac{1}{2} \left[ \left( \frac{\pi}{2} - 0 \right) - (0 - 0) \right] = \frac{1}{2} \times \frac{\pi}{2} = \frac{\pi}{4} $
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Calculate Final Value: Multiply the integral result by the constant factor $\frac{4}{\pi}$.
$ \text{Value} = \frac{4}{\pi} \times \frac{\pi}{4} $
$ \text{Value} = 1 $
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Rounding: Round the result to two decimal places.
The value is exactly 1. Rounded to two decimal places, it is 1.00.
This result aligns with the information that the correct value lies between 1 and 1, indicating the value is approximately 1.
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 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 toLet $C: x^2 + y^2 =9$ be the circle in $R^2$ oriented positively. Then $\frac{1}{\pi}\oint_C(3y-e^{\cos x})dx+(7x + \sqrt{y^4 +11})dy$ equals_____.
