Integral Simplification
First, simplify the expression inside the absolute value: $ \sin 3x + \sin 2x + \sin x $
Using the sum-to-product formula $ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) $, we combine $ \sin 3x $ and $ \sin x $: $ \sin 3x + \sin x = 2 \sin\left(\frac{3x+x}{2}\right) \cos\left(\frac{3x-x}{2}\right) = 2 \sin(2x) \cos(x) $
The expression becomes: $ (2 \sin(2x) \cos(x)) + \sin 2x = \sin 2x (2 \cos x + 1) $
Let $ f(x) = \sin 2x (2 \cos x + 1) $. The problem asks for the value of $ 6 \int_0^\pi |f(x)| dx $.
Integrand Sign Analysis
Determine the sign of $ f(x) = \sin 2x (2 \cos x + 1) $ in the interval $ [0, \pi] $.
- The term $ \sin 2x $ is zero at $ x = 0, \pi/2, \pi $.
- The term $ 2 \cos x + 1 $ is zero when $ \cos x = -1/2 $, which occurs at $ x = 2\pi/3 $ in the interval $ [0, \pi] $.
The interval $ [0, \pi] $ is divided into sub-intervals by these points: $ [0, \pi/2] $, $ [\pi/2, 2\pi/3] $, and $ [2\pi/3, \pi] $.
- Interval $ (0, \pi/2) $: $ \sin 2x > 0 $ and $ \cos x > 0 $, so $ 2 \cos x + 1 > 0 $. Thus, $ f(x) > 0 $.
- Interval $ (\pi/2, 2\pi/3) $: $ \sin 2x < 0 $ (since $ \pi/2 < x < \pi $) and $ \cos x $ is between $ 0 $ and $ -1/2 $. So $ 2 \cos x + 1 > 0 $. Thus, $ f(x) < 0 $.
- Interval $ (2\pi/3, \pi) $: $ \sin 2x < 0 $ and $ \cos x < -1/2 $. So $ 2 \cos x + 1 < 0 $. Thus, $ f(x) > 0 $.
Integral Splitting
Since the sign of $ f(x) $ changes, we split the integral according to the definition of absolute value $ |f(x)| $:
$ 6 \int_0^\pi |f(x)| dx = 6 \left[ \int_0^{\pi/2} f(x) dx + \int_{\pi/2}^{2\pi/3} (-f(x)) dx + \int_{2\pi/3}^{\pi} f(x) dx \right] $Indefinite Integral Calculation
Calculate the indefinite integral of the original expression: $ \int (\sin 3x + \sin 2x + \sin x) dx = -\frac{\cos 3x}{3} - \frac{\cos 2x}{2} - \cos x + C $
Let $ F(x) = -\frac{\cos 3x}{3} - \frac{\cos 2x}{2} - \cos x $.
Definite Integral Evaluation
Evaluate $ F(x) $ at the boundary points $ 0, \pi/2, 2\pi/3, \pi $:
- $ F(0) = -\frac{1}{3} - \frac{1}{2} - 1 = -\frac{11}{6} $
- $ F(\pi/2) = 0 - (-\frac{1}{2}) - 0 = \frac{1}{2} $
- $ F(2\pi/3) = -\frac{1}{3} - \frac{\cos(4\pi/3)}{2} - \cos(2\pi/3) = -\frac{1}{3} - \frac{-1/2}{2} - (-\frac{1}{2}) = -\frac{1}{3} + \frac{1}{4} + \frac{1}{2} = \frac{5}{12} $
- $ F(\pi) = -\frac{\cos(3\pi)}{3} - \frac{\cos(2\pi)}{2} - \cos(\pi) = - \frac{-1}{3} - \frac{1}{2} - (-1) = \frac{1}{3} - \frac{1}{2} + 1 = \frac{5}{6} $
Calculate the definite integrals over the sub-intervals:
- $ \int_0^{\pi/2} f(x) dx = F(\pi/2) - F(0) = \frac{1}{2} - (-\frac{11}{6}) = \frac{3}{6} + \frac{11}{6} = \frac{14}{6} = \frac{7}{3} $
- $ \int_{\pi/2}^{2\pi/3} f(x) dx = F(2\pi/3) - F(\pi/2) = \frac{5}{12} - \frac{1}{2} = \frac{5}{12} - \frac{6}{12} = -\frac{1}{12} $
- $ \int_{2\pi/3}^{\pi} f(x) dx = F(\pi) - F(2\pi/3) = \frac{5}{6} - \frac{5}{12} = \frac{10}{12} - \frac{5}{12} = \frac{5}{12} $
Final Calculation
Substitute the values back into the split integral expression: $ 6 \left[ \frac{7}{3} + -(-\frac{1}{12}) + \frac{5}{12} \right] = 6 \left[ \frac{7}{3} + \frac{1}{12} + \frac{5}{12} \right] $
Find a common denominator and sum the fractions: $ 6 \left[ \frac{7 \times 4}{12} + \frac{1}{12} + \frac{5}{12} \right] = 6 \left[ \frac{28 + 1 + 5}{12} \right] = 6 \left[ \frac{34}{12} \right] $
Simplify the final expression: $ 6 \times \frac{34}{12} = 6 \times \frac{17}{6} = 17 $
Similar Questions
- Let the line $x = -1$ divide the area of the region $\{(x, y) : 1 + x^2 \le y \le 3 - x\}$ in the ratio $m : n, \gcd(m, n) = 1$. Then $m + n$ is equal to
- Let $f(x) = [x]^2 - [x+3] - 3, x \in \mathbb{R}$, where $[\cdot]$ is the greatest integer function. Then
- If $\lim_{x \to 0} \frac{e^{(a-1)x} + 2 \cos bx + (c-2)e^{-x}}{x \cos x - \log_e(1+x)} = 2$, then $a^2 + b^2 + c^2$ is equal to :
Let $[\cdot]$ be the greatest integer function. If $\alpha = \int_{0}^{64} (x^{1/3} - [x^{1/3}]) dx$, then $\frac{1}{\pi} \int_{0}^{\alpha \pi} \left(\frac{\sin^2 \theta}{\sin^6 \theta + \cos^6 \theta}\right) d\theta$ is equal to ________.
- Let $y = y(x)$ be a differentiable function in the interval $(0, \infty)$ such that $y(1) = 2$, and $\lim_{t \rightarrow x} \left( \frac{t^2 y(x) - x^2 y(t)}{x - t} \right) = 3$ for each $x > 0$. Then $2y(2)$ is equal to
- If $f(x)$ satisfies the relation $f(x) = e^x + \int_0^1 (y + xe^x)f(y)dy$, then $e + f(0)$ is equal to _______.
- The value of $\int_{-\pi/6}^{\pi/6} \left( \frac{\pi + 4x^{11}}{1 - \sin(|x| + \pi/6)} \right) dx$ is equal to :
- Let $y = y(x)$ be the solution curve of the differential equation $(1 + x^2) dy + (y - \tan^{-1}x) dx = 0$, $y(0) = 1$. Then the value of $y(1)$ is :
- Let $y = y(x)$ be the solution of the differential equation $\sec x \frac{dy}{dx} - 2y = 2 + 3\sin x$, $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, $y(0) = -\frac{7}{4}$. Then $y\left(\frac{\pi}{6}\right)$ is equal to :
- If $\int_0^1 4 \cot^{-1}(1 - 2x + 4x^2) dx = a \tan^{-1}(2) - b \log_e(5)$, where $a, b \in \mathbf{N}$, then $(2a + b)$ is equal to __________.
Important Questions from Calculus
- Let the line $x = -1$ divide the area of the region $\{(x, y) : 1 + x^2 \le y \le 3 - x\}$ in the ratio $m : n, \gcd(m, n) = 1$. Then $m + n$ is equal to
- Let $f(x) = [x]^2 - [x+3] - 3, x \in \mathbb{R}$, where $[\cdot]$ is the greatest integer function. Then
- If $\lim_{x \to 0} \frac{e^{(a-1)x} + 2 \cos bx + (c-2)e^{-x}}{x \cos x - \log_e(1+x)} = 2$, then $a^2 + b^2 + c^2$ is equal to :
Let $[\cdot]$ be the greatest integer function. If $\alpha = \int_{0}^{64} (x^{1/3} - [x^{1/3}]) dx$, then $\frac{1}{\pi} \int_{0}^{\alpha \pi} \left(\frac{\sin^2 \theta}{\sin^6 \theta + \cos^6 \theta}\right) d\theta$ is equal to ________.
- Let $y = y(x)$ be a differentiable function in the interval $(0, \infty)$ such that $y(1) = 2$, and $\lim_{t \rightarrow x} \left( \frac{t^2 y(x) - x^2 y(t)}{x - t} \right) = 3$ for each $x > 0$. Then $2y(2)$ is equal to
