Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

\(\displaystyle \int_{-2}^{2}14x^3+5x^2dx\)

Solution:

\(\displaystyle \int14x^3+5x^2dx\)

\(\displaystyle \frac{14}{4}x^4+\frac{5}{3}x^3\Big|_{-2}^{2}\)

\(\displaystyle (\frac{7}{2}2^4 + \frac{5}{3}2^3)-(\frac{7}{2}(-2)^4+\frac{5}{3}(-2)^3)\)

\(\displaystyle (\frac{7}{2}\cdot16+\frac{5}{3}\cdot8)-(\frac{7}{2}\cdot16-\frac{5}{3}\cdot8)\)

\(\displaystyle (56+\frac{40}{3})-(56-\frac{40}{3}) = \frac{80}{3} = 26.67\)

When rounded, the area under the curve is \(\displaystyle 27\)

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

\(\displaystyle \int_{0}^{2}3x^2+e^xdx\)

Solution:

\(\displaystyle \int3x^2+e^xdx\)

\(\displaystyle \frac{3}{3}x^3+e^x\Big|_{0}^{2}\)

\(\displaystyle x^3+e^x\Big|_{0}^{2}\)

\(\displaystyle (2^3+e^2)-(e^0)\)

\(\displaystyle 8+e^2-1 = e^2 + 7 = 14.389\)

When rounded to the nearest integer, the area under the curve is \(\displaystyle 14\)

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

\(\displaystyle \int_{0}^{2}7x+e^{2x}dx\)

Solution:

\(\displaystyle \int7x+e^{2x}dx\)

\(\displaystyle \frac{7}{2}x^2+\frac{1}{2}e^{2x}\Big|_{0}^{2}\)

\(\displaystyle (\frac{7}{2}\cdot2^2+\frac{1}{2}e^{2\cdot2})-(\frac{1}{2})\)

\(\displaystyle 14+\frac{1}{2}e^4-\frac{1}{2} = \frac{27}{2}+\frac{1}{2}e^4 = 40.799\)

When rounded, the area under the curve is \(\displaystyle 41\)

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

\(\displaystyle \int_{-2}^{2}|4x+x^2|dx\)

Solution:

This function is negative from x=[-2,0], and positve everywhere else. Split this integral up into 2 pieces, multiplying x=[-2,0] region by -1, and sum everything up.

1st Piece:

\(\displaystyle \int_{-2}^{0}-4x-x^2dx\)

\(\displaystyle -2x^2-\frac{1}{3}x^3\Big|_{-2}^{0}\)

\(\displaystyle 0-(-2\cdot4-\frac{1}{3}\cdot(-8)) = -(-8+\frac{8}{3}) = 8 - \frac{8}{3} = \frac{16}{3}\)

2nd piece:

\(\displaystyle \int_{0}^{2}4x+x^2dx\)

\(\displaystyle 2x^2+\frac{1}{3}x^3\Big|_{0}^{2}\)

\(\displaystyle 2\cdot4+\frac{8}{3} = 8+\frac{8}{3} = \frac{32}{3}\)

Sum:

\(\displaystyle \frac{16}{3} + \frac{32}{3} = \frac{48}{3} = 16\)

The area under the curve is \(\displaystyle 16\)

inding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

\(\displaystyle \int_{-2}^{2}|-6x+2x^2|dx\)

Solution:

This function is negative from x=[0,2], and positve everywhere else. Split this integral up into 2 pieces, multiplying x=[0,2] region by -1, and sum everything up.

1st piece:

\(\displaystyle \int_{-3}^{0}|-6x+2x^2|dx\)

\(\displaystyle -3x^2+\frac{2}{3}x^3\Big|_{-3}^{0}\)

\(\displaystyle 0-(-3\cdot(-3)^2+\frac{2}{3}\cdot(-3)^3)\)

\(\displaystyle 9+\frac{2}{3}\cdot27 = 9+18=27\)

2nd piece:

\(\displaystyle \int_{0}^{2}|-6x+2x^2|dx\)

\(\displaystyle 3x^2-\frac{2}{3}x^3\Big|_{0}^{2}\)

\(\displaystyle 3*4-\frac{2}{3}\cdot8 = 12-\frac{16}{3}\)

sum:

\(\displaystyle 27+12-\frac{16}{3} = \frac{155}{3} = 51.667\)

When rounded to the nearest integer, the area under the curve is \(\displaystyle 52\)

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

\(\displaystyle \int_{-1}^{1}|-3x+2x^2+x^3|dx\)

Solution:

This function is negative from \(\displaystyle x=[0,1]\), and positve everywhere else. Split this integral up into 2 pieces, multiplying \(\displaystyle x=[0,1]\) region by \(\displaystyle -1\), and sum everything up.

First piece:

\(\displaystyle \int_{-1}^{0}|-3x+2x^2+x^3|dx\)

\(\displaystyle -\frac{3}{2}x^2+\frac{2}{3}x^3+\frac{1}{4}x^4\Big|_{-1}^{0}\)

\(\displaystyle 0-(-\frac{3}{2}-\frac{2}{3}+\frac{1}{4}) = \frac{3}{2}+\frac{2}{3}-\frac{1}{4} = \frac{23}{12}\)

Second piece:

\(\displaystyle \int_{0}^{1}|-3x+2x^2+x^3|dx\)

\(\displaystyle \frac{3}{2}x^2-\frac{2}{3}x^3-\frac{1}{4}x^4\Big|_{0}^{1}\)

\(\displaystyle \frac{3}{2}-\frac{2}{3}-\frac{1}{4} = \frac{7}{12}\)

Sum:

\(\displaystyle \frac{23}{12}+\frac{7}{12} = \frac{30}{12} = \frac{5}{2}=2.5\)

When rounded to the nearest integer, the area under the curve is \(\displaystyle 3\)

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

\(\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}sin(x)+1dx\)

Solution:

\(\displaystyle \int sin(x)+1dx\)

\(\displaystyle x-cos(x)\Big|_{\frac{\pi}{4}}^{\frac{\pi}{2}}\)

\(\displaystyle ({\frac{\pi}{2}}-cos({\frac{\pi}{2}}))-({\frac{\pi}{4}} - cos({\frac{\pi}{4}}))\)

\(\displaystyle \frac{\pi}{4} + \frac{1}{\sqrt2} = 1.4925\)

Rounded to the nearest integer, the area under the curve is \(\displaystyle 1\)

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

\(\displaystyle \int_{0}^{\pi}2sin(x)+xdx\)

Solution:

\(\displaystyle \int2sin(x)+xdx\)

\(\displaystyle \frac{1}{2}x^2-2cos(x)\Big|_{0}^{\pi}\)

\(\displaystyle \frac{1}{2}\pi^2-2\cdot cos(\pi)-(-2cos(0)) = \frac{1}{2}\pi^2-2\cdot cos(\pi) + 2cos(0)\)

\(\displaystyle \frac{1}{2}\pi^2-2\cdot-1 + 2 = \frac{1}{2}\pi^2 + 4\)

The area under the curve is \(\displaystyle \frac{1}{2}\pi^2 + 4\)

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

\(\displaystyle \int_{-2}^{2}|3x^2-4x|dx\)

Solution:

This function is negative from \(\displaystyle x=[0,\frac{4}{3}],\) and positve everywhere else. Split this integral up into 3 pieces, multiplying x=[0,\frac{4}{3}] region by -1, and sum everything up.

1st piece:

\(\displaystyle \int_{-2}^{0}3x^2-4xdx\)

\(\displaystyle 3\cdot\frac{1}{3}x^3-\frac{4}{2}x^2\Big|_{-2}^{0}\)

\(\displaystyle x^3-2x^2\Big|_{-2}^{0}\)

\(\displaystyle 0-((-2)^3-2\cdot4) = 8+8 = 16\)

2nd piece:

\(\displaystyle \int_{0}^{\frac{4}{3}}|3x^2-4x|dx\)

\(\displaystyle 2x^2 - x^3\Big|_{0}^{\frac{4}{3}}\)

\(\displaystyle 2\cdot(\frac{4}{3})^2 - (\frac{4}{3})^3 = \frac{32}{9}-\frac{64}{27} = \frac{32}{27}\)

3rd piece:

\(\displaystyle \int_{\frac{4}{3}}^{2}|3x^2-4x|dx\)

\(\displaystyle x^3 - 2x^2\Big|_{\frac{4}{3}}^{2}\)

\(\displaystyle (8-2\cdot4)-((\frac{4}{3})^3-2\cdot(\frac{4}{3})^2) = \frac{32}{9}-\frac{64}{27} = \frac{32}{27}\)

Sum:

\(\displaystyle 16+\frac{32}{27} + \frac{32}{27} = \frac{496}{27} = 18.37\)

When rounded to the nearest integer, the area under the curve is \(\displaystyle 18\)

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

\(\displaystyle \int_{1}^{4} (x+6)^2-36dx\)

Solution:

First, simplify the function and then evaluate the integral.

1. Simplify the function

\(\displaystyle (x+6)^2-36\)

\(\displaystyle x^2+36+12x-36\)

\(\displaystyle x^2+12x\)

2. Evaluate the integral

\(\displaystyle \int_{1}^{4} x^2+12xdx\)

\(\displaystyle \frac{1}{3}x^3+6x^2\Big|_{1}^{4}\)

\(\displaystyle \frac{64}{3}+6\cdot16-(\frac{1}{3}+6) = 111\)

The area under the curve is \(\displaystyle 111\)