Question

Let $f(\theta) = \sin \theta + \int_{-\pi/2}^{\pi/2} (\sin \theta + t\cos \theta)f(t)dt$. Then the value of

$\left| \int_{0}^{\pi/2} f(\theta)d\theta \right|$ is ___.

Answer 1

1

Answer 2

The given integral equation is: $f(\theta) = \sin \theta + \int_{-\pi/2}^{\pi/2} (\sin \theta + t\cos \theta)f(t)dt$

First, separate the terms inside the integral: $f(\theta) = \sin \theta + \int_{-\pi/2}^{\pi/2} \sin \theta f(t)dt + \int_{-\pi/2}^{\pi/2} t\cos \theta f(t)dt$

Since $\sin \theta$ and $\cos \theta$ are constants with respect to the integration variable $t$, we can take them out of the integral: $f(\theta) = \sin \theta + \sin \theta \int_{-\pi/2}^{\pi/2} f(t)dt + \cos \theta \int_{-\pi/2}^{\pi/2} t f(t)dt$

Let $A = \int_{-\pi/2}^{\pi/2} f(t)dt$ and $B = \int_{-\pi/2}^{\pi/2} t f(t)dt$. Then the expression for $f(\theta)$ becomes: $f(\theta) = \sin \theta + A \sin \theta + B \cos \theta$ $f(\theta) = (1+A) \sin \theta + B \cos \theta$

Now, we need to determine the values of A and B.

Substitute $f(t)$ back into the definition of A: $A = \int_{-\pi/2}^{\pi/2} ((1+A) \sin t + B \cos t) dt$ $A = (1+A) \int_{-\pi/2}^{\pi/2} \sin t dt + B \int_{-\pi/2}^{\pi/2} \cos t dt$

Evaluate the integrals: $\int_{-\pi/2}^{\pi/2} \sin t dt = [-\cos t]_{-\pi/2}^{\pi/2} = -\cos(\pi/2) - (-\cos(-\pi/2)) = 0 - 0 = 0$. $\int_{-\pi/2}^{\pi/2} \cos t dt = [\sin t]_{-\pi/2}^{\pi/2} = \sin(\pi/2) - \sin(-\pi/2) = 1 - (-1) = 2$.

Substitute these values back into the equation for A: $A = (1+A)(0) + B(2)$ $A = 2B$ (Equation 1)

Next, substitute $f(t)$ back into the definition of B: $B = \int_{-\pi/2}^{\pi/2} t ((1+A) \sin t + B \cos t) dt$ $B = (1+A) \int_{-\pi/2}^{\pi/2} t \sin t dt + B \int_{-\pi/2}^{\pi/2} t \cos t dt$

Evaluate the integrals: For $\int_{-\pi/2}^{\pi/2} t \sin t dt$: The integrand $g(t) = t \sin t$ is an even function ($g(-t) = (-t)\sin(-t) = t \sin t = g(t)$). So, $\int_{-\pi/2}^{\pi/2} t \sin t dt = 2 \int_{0}^{\pi/2} t \sin t dt$. Using integration by parts ($\int u dv = uv - \int v du$ with $u=t, dv=\sin t dt \implies du=dt, v=-\cos t$): $\int_{0}^{\pi/2} t \sin t dt = [-t \cos t - \int (-\cos t) dt]_{0}^{\pi/2} = [-t \cos t + \sin t]_{0}^{\pi/2}$ $= (-\pi/2 \cos(\pi/2) + \sin(\pi/2)) - (0 \cos 0 + \sin 0)$ $= (0 + 1) - (0 + 0) = 1$. Thus, $\int_{-\pi/2}^{\pi/2} t \sin t dt = 2(1) = 2$.

For $\int_{-\pi/2}^{\pi/2} t \cos t dt$: The integrand $h(t) = t \cos t$ is an odd function ($h(-t) = (-t)\cos(-t) = -t \cos t = -h(t)$). For an odd function over a symmetric interval $[-a, a]$, the integral is 0. So, $\int_{-\pi/2}^{\pi/2} t \cos t dt = 0$.

Substitute these values back into the equation for B: $B = (1+A)(2) + B(0)$ $B = 2(1+A)$ (Equation 2)

Now we have a system of two linear equations for A and B:

1. $A = 2B$
2. $B = 2(1+A)$

Substitute Equation 1 into Equation 2: $B = 2(1 + 2B)$ $B = 2 + 4B$ $3B = -2$ $B = -2/3$

Now find A using Equation 1: $A = 2B = 2(-2/3) = -4/3$

So, $A = -4/3$ and $B = -2/3$.

Now we can write the explicit form of $f(\theta)$: $f(\theta) = (1+A) \sin \theta + B \cos \theta$ $f(\theta) = (1 - 4/3) \sin \theta + (-2/3) \cos \theta$ $f(\theta) = (-1/3) \sin \theta - (2/3) \cos \theta$

Finally, we need to calculate $\left| \int_{0}^{\pi/2} f(\theta)d\theta \right|$: $\int_{0}^{\pi/2} f(\theta)d\theta = \int_{0}^{\pi/2} \left( -\frac{1}{3} \sin \theta - \frac{2}{3} \cos \theta \right) d\theta$ $= -\frac{1}{3} \int_{0}^{\pi/2} \sin \theta d\theta - \frac{2}{3} \int_{0}^{\pi/2} \cos \theta d\theta$

Evaluate these definite integrals: $\int_{0}^{\pi/2} \sin \theta d\theta = [-\cos \theta]_{0}^{\pi/2} = -\cos(\pi/2) - (-\cos(0)) = 0 - (-1) = 1$. $\int_{0}^{\pi/2} \cos \theta d\theta = [\sin \theta]_{0}^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1$.

Substitute these values back: $\int_{0}^{\pi/2} f(\theta)d\theta = -\frac{1}{3}(1) - \frac{2}{3}(1)$ $= -\frac{1}{3} - \frac{2}{3} = -\frac{3}{3} = -1$.

The value of $\left| \int_{0}^{\pi/2} f(\theta)d\theta \right|$ is $|-1| = 1$.

The final answer is $\boxed{1}$.