Given that the region is enclosed by the x-axis and this curve:
[tex]y=-x^2-3x+4[/tex]
You can graph the function using a Graphic Tool:
Noice that the area region you must calculate is:
Notice that it goes from:
[tex]x=-4[/tex]
To:
[tex]x=1[/tex]
Therefore, you can set up that:
[tex]Area=\int_{-4}^1(x^2-3x+4)-(0)dx[/tex]
In order to solve the Definite Integral, you need to:
- Apply these Integration Rules:
[tex]\int x^ndx=\frac{x^{n+1}}{n+1}+C[/tex][tex]\int kf(x)dx=k\int f(x)dx[/tex]
Then, you get:
[tex]=(\frac{x^3}{3}-\frac{3x^2}{2}+4)|^1_{-4}[/tex]
- Evaluate:
[tex]=(\frac{1^3}{3}-\frac{3(1)^2}{2}+4)-(\frac{(-4)^3}{3}-\frac{3(-4)^2}{2}+4)[/tex][tex]Area\approx64.17[/tex]
Hence, the answer is:
[tex]Area\approx64.17[/tex]
