Question

question 6 (1 point)
the graph of ( y = f ( t ) ) is given below:

the definite integral ( int _ { 0 } ^ { 6 } f ( t ) d t ) equals
(your answer should be a number.)
view hint for question 6

Explanation:

Step1: Calculate the area of the first triangle

The first triangle has base \( b_1=2\) (from \(t = 0\) to \(t = 2\)) and height \(h_1 = 1\). The area of a triangle is \(A=\frac{1}{2}\times base\times height\). So \(A_1=\frac{1}{2}\times2\times1 = 1\) (but since it's below the \(t\)-axis in the first - part (wait, no, actually, re - evaluate: from \(t = 0\) to \(t=2\), the function \(y = f(t)\) forms a triangle. Wait, no, looking at the graph: from \(t = 0\) to \(t = 2\), the function \(y=f(t)\) has a "valley". The base of the first (lower) triangle: from \(t = 0\) to \(t = 2\), base \(b = 2\), height \(h = 1\) (from \(y = 0\) to \(y=- 1\)). Area \(A_{1}=\frac{1}{2}\times2\times(-1)=-1\). From \(t = 2\) to \(t = 6\), we have two parts: a trapezoid - like (but actually, from \(t = 2\) to \(t = 6\), first a line from \((2,-1)\) to \((6,3)\). The area under the curve from \(t = 2\) to \(t = 6\) can be calculated as the area of a trapezoid. The formula for the area of a trapezoid is \(A=\frac{(a + b)h}{2}\), where \(a\) and \(b\) are the parallel sides and \(h\) is the height (the difference in the \(t\) - values). Here, \(a=-1\), \(b = 3\), and \(h=6 - 2=4\). So \(A_{2}=\frac{(-1 + 3)\times4}{2}=4\)

Step2: Sum the areas

The definite integral \(\int_{0}^{6}f(t)dt\) is the sum of the areas of the regions between the curve \(y = f(t)\) and the \(t\) - axis. \(\int_{0}^{6}f(t)dt=A_1+A_2\). Substitute \(A_1=-1\) and \(A_2 = 4\)

Answer:

\(3\)