# Problem with solution Simplify 6 Integrals with the Delta Function Level 3 (with higher mathematics)
Level 3 requires the basics of vector calculus, differential and integral calculus. Suitable for undergraduates and high school students.

Calculate the following integrals containing a delta function $$\delta(x)$$:

1. Integral 1
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2. Integral 2
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3. Integral 3
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4. Integral 4
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5. Integral 5
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6. Integral 6
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Solution tips
• First, check that the delta function is within the integration limits.
• If the delta function lies within the integration limits, then evaluate $$f(x)$$ at the position $$p$$ of the delta function:$$\int f(x) \, \delta(\cdot) \, \text{d}x ~=~ f(p)$$
Solution for (a)

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}

Solution for (b)

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}

Solution for (c)

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}

Solution for (d)

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}

Solution for (e)

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}

Solution for (f)

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