Question

using geometry, determine $int_{-4}^{4}(2 - |t|)dt$. attempt 1: 2 attempts remaining. submit answer next item

Explanation:

Step1: Analyze the function's symmetry

The function \(y = 2-\vert t\vert\) is an even - function since \(f(-t)=2-\vert - t\vert=2 - \vert t\vert=f(t)\). For an even function \(y = f(t)\), \(\int_{-a}^{a}f(t)dt = 2\int_{0}^{a}f(t)dt\). Here \(a = 4\), so \(\int_{-4}^{4}(2-\vert t\vert)dt=2\int_{0}^{4}(2 - t)dt\) (when \(t\geq0\), \(\vert t\vert=t\)).

Step2: Use the integral rules

We know that \(\int_{0}^{4}(2 - t)dt=\int_{0}^{4}2dt-\int_{0}^{4}tdt\). According to the power - rule of integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\) and \(\int kdx=kx + C\) (where \(k\) is a constant). So \(\int_{0}^{4}2dt=2t\big|_{0}^{4}=2\times4-2\times0 = 8\) and \(\int_{0}^{4}tdt=\frac{t^{2}}{2}\big|_{0}^{4}=\frac{4^{2}}{2}-\frac{0^{2}}{2}=8\).

Step3: Calculate the result

Then \(2\int_{0}^{4}(2 - t)dt=2(8 - 8)=0\).

Answer:

0