Problem with solution Simplify 6 Integrals with the Delta Function

We want to calculate the following integral:$$ \int^5_1 \left( 2x^2 - x + 1 \right) \, \delta(x-3) \, \text{d}x $$Here \(f(x) = 2x^2 - x + 1 \). And the position of the delta function is at \(x=3\).

First, we ask ourselves if the position of \(\delta(x-3)\) lies in the integration interval [1, 5]. It does. Therefore, the integral is not necessarily zero and we must evaluate the function \(f(x)\) at the position \(x=3\) where the delta function is found:\begin{align} \int^5_1 \left( 2x^2 - x + 1 \right) \, \delta(x-3) \, \text{d}x ~&=~ 2\cdot 3^2 - 3 + 1 \\ ~&=~ 16 \end{align}

We want to calculate the following integral:$$ \int^{1.5}_0 x^3 \, \delta(x+2) \, \text{d}x $$Here \(f(x) = x^3 \). And the delta function is shifted to the negative and is at \(x=-2\).

First, we ask ourselves if the position of \(\delta(x+2)\) lies in the integration interval [0, 1.5]. It does. Therefore, the integral is not necessarily zero and we must evaluate the function \(f(x)\) at the position \(x=-2\) where the delta function is found:\begin{align} \int^{1.5}_0 x^3 \, \delta(x+2) \, \text{d}x ~&=~ (-2)^3 \\ ~&=~ -8 \end{align}

We want to calculate the following integral:$$ \int^4_0 \cos(x) \, \delta(x-\pi) \, \text{d}x $$Here \(f(x) = \cos(x) \). And the position of the delta function is at \(x=\pi\).

First, we ask ourselves if the position of \(\delta(x-\pi)\) lies in the integration interval [0, 4]. It does. Therefore, the integral is not necessarily zero and we must evaluate the function \(f(x)\) at the position \(x=\pi\) where the delta function is found:\begin{align} \int^4_0 \cos(x) \, \delta(x-\pi) \, \text{d}x ~&=~ \cos(\pi) \\ ~&=~ -1 \end{align}

We want to calculate the following integral:$$ \int \ln(x+3) \, \delta(x+1) \, \text{d}x $$Here \(f(x) = \ln(x+3) \). And the position of the delta function is at \(x=-1\).

First, we ask ourselves if the position of \(\delta(x+1)\) lies in the integration interval \([-\infty, \infty]\). It does. Therefore, the integral is not necessarily zero and we must evaluate the function \(f(x)\) at the position \(x=-1\) where the delta function can be found:\begin{align} \int \ln(x+3) \, \delta(x+1) \, \text{d}x ~&=~ \ln(-1 + 3) \\ ~&=~ 0.693... \end{align}

We want to calculate the following integral:$$ \int^2_{-3} \left( 6x + 2 \right) \, \delta(3x) \, \text{d}x $$Here \(f(x) = \left( 6x + 2 \right) \). And the position of the delta function is at \(x=0\). The scaling factor is \( |k| = 3 \).

First we question whether the position of \(\delta(3x)\) lies in the integration interval \([-3, 2]\). It does. Therefore, the integral is not necessarily zero and we need to evaluate the function \(f(x)\) at the position \(x=0\) and multiply it by the factor \(1/|k|\):\begin{align} \int^2_{-3} \left( 6x + 2 \right) \, \delta(3x) \, \text{d}x ~&=~ \frac{1}{3}\, \left( 6\cdot 0 + 2 \right) \\ ~&=~ \frac{2}{3} \end{align}