Question

Pre calculus
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Answer 1

`\displaystyle (75)/(2)` or
`37.5`

Explanation:

We can answer this problem geometrically:

`\displaystyle \int^6_(-4)f(x)\,dx=\int^1_(-4)f(x)\,dx+\int^3_1f(x)\,dx+\int^6_3f(x)\,dx\\\\\int^6_(-4)f(x)\,dx=(5*5)+(1)/(2)(2*5)+(1)/(2)(3*5)\\\\\int^6_(-4)f(x)\,dx=25+5+7.5\\\\\int^6_(-4)f(x)\,dx=37.5=(75)/(2)`

Notice that we found the area of the rectangular region between -4 and 1, and then the two triangular areas from 1 to 3 and 3 to 6. We then found the sum of these areas to get the total area under the curve of f(x) from -4 to 6.

Answer 2

`(75)/(2)`

Explanation:

The value of a definite integral represents the area between the x-axis and the graph of the function you’re integrating between two limits.

`\boxed{\begin{minipage}{8.5 cm}\underline{De\:\!finite integration}\\\\$\displaystyle \int^b_a f(x)\:\:\text{d}x$\\\\\\where $a$ is the lower limit and $b$ is the upper limit.\\\end{minipage}}`

The given definite integral is:

`\displaystyle \int^6_(-4) f(x)\; \;\text{d}x`

This means we need to find the area between the x-axis and the function between the limits x = -4 and x = 6.

Notice that the function touches the x-axis at x = 3.

Therefore, we can separate the integral into two areas and add them together:

`\displaystyle \int^6_(-4) f(x)\; \;\text{d}x=\int^3_(-4) f(x)\; \;\text{d}x+\int^6_(3) f(x)\; \;\text{d}x`

The area between the x-axis and the function between the limits x = -4 and x = 3 is a trapezoid with bases of 5 and 7 units, and a height of 5 units.

The area between the x-axis and the function between the limits x = 3 and x = 6 is a triangle with base of 3 units and height of 5 units.

Using the formulas for the area of a trapezoid and the area of a triangle, the definite integral can be calculated as follows:

`\begin{aligned}\displaystyle \int^6_(-4) f(x)\; \;\text{d}x & =\int^3_(-4) f(x)\; \;\text{d}x+\int^6_(3) f(x)\; \;\text{d}x\\\\& =(1)/(2)(5+7)(5)+(1)/(2)(3)(5)\\\\& =30+(15)/(2)\\\\& =(75)/(2)\end{aligned}`