In order to find the area underneath a curve for a specific range of values, we need to set up a definite integral expression. To do this, we plug our specific information into the following skeleton:

\(\displaystyle \int_{a}^{b}f(x) dx\)

We plug in the equation of the curve in for \(\displaystyle f(x)\), the smaller of our x-values for \(\displaystyle a\), and the larger of our x-values for \(\displaystyle b\). For this problem, our set-up looks like this:

\(\displaystyle \int_{0}^{1}(45x^{2}+10x+14)dx\)

Next, we integrate our expression, ignoring our \(\displaystyle a\) and \(\displaystyle b\) values for now. We get:

\(\displaystyle F(x)= 15x^{3}+5x^{2}+14x\)

Now, to find the area under the part of the curve we're looking for, we plug our \(\displaystyle a\) and \(\displaystyle b\) values into our integrated expression and find the difference, using the following skeleton:

\(\displaystyle area = F(b)-F(a)\)

In this problem, this looks like:

\(\displaystyle area = F(1)-F(0)\)

Which plugged into our integrated expression is:

\(\displaystyle area = [15(1)^{3}+5(1)^{2}+14(1)] - [15(0)^{3}+5(0)^{2}+14(0)]\)

\(\displaystyle area = [15+5+14] - [0+0+0]\)

\(\displaystyle area = [34] - [0]\)

With our final answer being:

\(\displaystyle area = 34\)

In order to find the area underneath a curve for a specific range of values, we need to set up a definite integral expression. To do this, we plug our specific information into the following skeleton:

\(\displaystyle \int_{a}^{b}f(x) dx\)

We plug in the equation of the curve in for \(\displaystyle f(x)\), the smaller of our x-values for \(\displaystyle a\), and the larger of our x-values for \(\displaystyle b\). For this problem, our set-up looks like this:

\(\displaystyle \int_{0}^{3}(42x^{6}+\frac{4}{7}x^{3}+76x)dx\)

Next, we integrate our expression, ignoring our \(\displaystyle a\) and \(\displaystyle b\) values for now. We get:

\(\displaystyle F(x)= 6x^{7}+\frac{1}{7}x^{4}+38x^{2}\)

Now, to find the area under the part of the curve we're looking for, we plug our \(\displaystyle a\) and \(\displaystyle b\) values into our integrated expression and find the difference, using the following skeleton:

\(\displaystyle area = F(b)-F(a)\)

In this problem, this looks like:

\(\displaystyle area = F(3)-F(0)\)

Which plugged into our integrated expression is:

\(\displaystyle area = [6(3)^{7}+\frac{1}{7}(3)^{4}+38(3)^{2}] - [6(0)^{7}+\frac{1}{7}(0)^{4}+38(0)^{2}]\)

\(\displaystyle area = [6(2187)+\frac{1}{7}(81)+38(9)] - [6(0)+\frac{1}{7}(0)+38(0)]\)

\(\displaystyle area = [13122+\frac{81}{7}+342)] - [0+0+0]\)

\(\displaystyle area = [\frac{91854}{7}+\frac{81}{7}+\frac{2394}{7})] - [0]\)

\(\displaystyle area = \frac{94329}{7}\)

With our final answer being:

\(\displaystyle area = \frac{94329}{7}\)

In order to find the area underneath a curve for a specific range of values, we need to set up a definite integral expression. To do this, we plug our specific information into the following skeleton:

\(\displaystyle \int_{a}^{b}f(x) dx\)

We plug in the equation of the curve in for \(\displaystyle f(x)\), the smaller of our x-values for \(\displaystyle a\), and the larger of our x-values for \(\displaystyle b\). For this problem, our set-up looks like this:

\(\displaystyle \int_{1}^{3}(12x^{2}+6x+12)dx\)

Next, we integrate our expression, ignoring our \(\displaystyle a\) and \(\displaystyle b\) values for now. We get:

\(\displaystyle F(x)= 4x^{3}+3x^{2}+12x\)

Now, to find the area under the part of the curve we're looking for, we plug our \(\displaystyle a\) and \(\displaystyle b\) values into our integrated expression and find the difference, using the following skeleton:

\(\displaystyle area = F(b)-F(a)\)

In this problem, this looks like:

\(\displaystyle area = F(3)-F(1)\)

Which plugged into our integrated expression is:

\(\displaystyle area = [4(3)^{3}+3(3)^{2}+12(3)] - [4(1)^{3}+3(1)^{2}+12(1)]\)

\(\displaystyle area = [4(27)+3(9)+12(3)] - [4(1)+3(1)+12(1)]\)

\(\displaystyle area = [108+27+36] - [4+3+12]\)

\(\displaystyle area = [171] - [19]\)

With our final answer being:

\(\displaystyle area = 152\)

In order to find the area underneath a curve for a specific range of values, we need to set up a definite integral expression. To do this, we plug our specific information into the following skeleton:

\(\displaystyle \int_{a}^{b}f(x) dx\)

We plug in the equation of the curve in for \(\displaystyle f(x)\), the smaller of our x-values for \(\displaystyle a\), and the larger of our x-values for \(\displaystyle b\). For this problem, our set-up looks like this:

\(\displaystyle \int_{-1}^{2}(12x^{3}+12x^{2}+ 12x+12 )dx\)

Next, we integrate our expression, ignoring our \(\displaystyle a\) and \(\displaystyle b\) values for now. We get:

\(\displaystyle F(x)= 3x^{4}+4x^{3}+6x^{2}+12x\)

Now, to find the area under the part of the curve we're looking for, we plug our \(\displaystyle a\) and \(\displaystyle b\) values into our integrated expression and find the difference, using the following skeleton:

\(\displaystyle area = F(b)-F(a)\)

In this problem, this looks like:

\(\displaystyle area = F(2)-F(-1)\)

Which plugged into our integrated expression is:

\(\displaystyle area = [3(2)^{4}+4(2)^{3}+6(2)^{2}+12(2)] - [3(-1)^{4}+4(-1)^{3}+6(-1)^{2}+12(-1)]\)

\(\displaystyle area = [3(16)+4(8)+6(4)+12(2)] - [3(1)+4(-1)+6(1)+12(-1)]\)

\(\displaystyle area = [48+32+24+24] - [3-4+6-12]\)

\(\displaystyle area = [128] - [-7]\)

With our final answer being:

\(\displaystyle area = 135\)

In order to find the area underneath a curve for a specific range of values, we need to set up a definite integral expression. To do this, we plug our specific information into the following skeleton:

\(\displaystyle \int_{a}^{b}f(x) dx\)

We plug in the equation of the curve in for \(\displaystyle f(x)\), the smaller of our x-values for \(\displaystyle a\), and the larger of our x-values for \(\displaystyle b\). For this problem, our set-up looks like this:

\(\displaystyle \int_{-2}^{2}(9x^{2}+14x+5)dx\)

Next, we integrate our expression, ignoring our \(\displaystyle a\) and \(\displaystyle b\) values for now. We get:

\(\displaystyle F(x)= 3x^{3}+7x^{2}+5x\)

Now, to find the area under the part of the curve we're looking for, we plug our \(\displaystyle a\) and \(\displaystyle b\) values into our integrated expression and find the difference, using the following skeleton:

\(\displaystyle area = F(b)-F(a)\)

In this problem, this looks like:

\(\displaystyle area = F(2)-F(-2)\)

Which plugged into our integrated expression is:

\(\displaystyle area = [3(2)^{3}+7(2)^{2}+5(2)] - [3(-2)^{3}+7(-2)^{2}+5(-2)]\)

\(\displaystyle area = [3(8)+7(4)+5(2)] - [3(-8)+7(4)+5(-2)]\)

\(\displaystyle area = [24+28+10] - [-24+28-10]\)

\(\displaystyle area = [62] - [-6]\)

With our final answer being:

\(\displaystyle area = 68\)

In order to find the area underneath a curve for a specific range of values, we need to set up a definite integral expression. To do this, we plug our specific information into the following skeleton:

\(\displaystyle \int_{a}^{b}f(x) dx\)

We plug in the equation of the curve in for \(\displaystyle f(x)\), the smaller of our x-values for \(\displaystyle a\), and the larger of our x-values for \(\displaystyle b\). For this problem, our set-up looks like this:

\(\displaystyle \int_{3}^{10}(7x+3)dx\)

Next, we integrate our expression, ignoring our \(\displaystyle a\) and \(\displaystyle b\) values for now. We get:

\(\displaystyle F(x)= \frac{17}{2}x^{2}+3x\)

Now, to find the area under the part of the curve we're looking for, we plug our \(\displaystyle a\) and \(\displaystyle b\) values into our integrated expression and find the difference, using the following skeleton:

\(\displaystyle area = F(b)-F(a)\)

In this problem, this looks like:

\(\displaystyle area = F(10)-F(3)\)

Which plugged into our integrated expression is:

\(\displaystyle area = [\frac{17}{2}(10)^{2}+3(10)] - [\frac{17}{2}(3)^{2}+3(3)]\)

\(\displaystyle area = [\frac{17}{2}(100)+3(10)] - [\frac{17}{2}(9)+3(3)]\)

\(\displaystyle area = [\frac{1700}{2}+30] - [\frac{153}{2}+9]\)

\(\displaystyle area = [\frac{1700}{2}+\frac{60}{2}] - [\frac{153}{2}+\frac{18}{2}]\)

\(\displaystyle area = [\frac{1760}{2}] - [\frac{171}{2}]\)

With our final answer being:

\(\displaystyle area = \frac{1589}{2}\)

In order to find the area underneath a curve for a specific range of values, we need to set up a definite integral expression. To do this, we plug our specific information into the following skeleton:

\(\displaystyle \int_{a}^{b}f(x) dx\)

We plug in the equation of the curve in for \(\displaystyle f(x)\), the smaller of our x-values for \(\displaystyle a\), and the larger of our x-values for \(\displaystyle b\). For this problem, our set-up looks like this:

\(\displaystyle \int_{1}^{2}(100x^{3}+6x)dx\)

Next, we integrate our expression, ignoring our \(\displaystyle a\) and \(\displaystyle b\) values for now. We get:

\(\displaystyle F(x)= 25x^{4}+3x^{2}\)

Now, to find the area under the part of the curve we're looking for, we plug our \(\displaystyle a\) and \(\displaystyle b\) values into our integrated expression and find the difference, using the following skeleton:

\(\displaystyle area = F(b)-F(a)\)

In this problem, this looks like:

\(\displaystyle area = F(1)-F(2)\)

Which plugged into our integrated expression is:

\(\displaystyle area = [25(2)^{4}+3(2)^{2}] - [25(1)^{4}+3(1)^{2})]\)

\(\displaystyle area = [25(16)+3(4)] - [25(1)+3(1)]\)

\(\displaystyle area = [400+12] - [25+3]\)

\(\displaystyle area = [412] - [28]\)

With our final answer being:

\(\displaystyle area = 384\)

To find the area bounded by the two curves, we must integrate:

\(\displaystyle \int_{a}^{b}[f(x)-g(x)]dx\), where f(x) is the top function and g(x) is the bottom function.

Using the above formula, we get

\(\displaystyle \int_{0}^{2}(2x+2-x^2)dx=2^2+2(2)-\frac{2^3}{3}-(0)=\frac{16}{3}\)

The integration was perfomed using the following rule:

\(\displaystyle \int x^n dx=\frac{x^{n+1}}{n+1}+C\)

The definite integration was performed by taking the result of the integral and evaluating it at the upper limit, and then subtracting the evaluation at the lower limit.

To determine the function that describes the area between two functions, we must integrate over the difference between the upper and lower functions:

\(\displaystyle \int ((x-5)^2-20-x^{\frac{1}{2}})dx\)

For ease, we can split the integral into three integrals:

\(\displaystyle \int (x-5)^2dx -\int 20 dx-\int x^{\frac{1}{2}}dx\)

For the first integral, we must make the following substitution:

\(\displaystyle u=x-5, du=dx\)

Rewriting the integral and integrating, we get

\(\displaystyle \int u^2=\frac{e^3}{3}+C\)

which was found using the following rule:

\(\displaystyle \int x^ndx=\frac{x^{n+1}}{n+1}+C\)

The second integral is equal to

\(\displaystyle \int 20 dx=20x+C\)

and was found using the following rule:

\(\displaystyle \int dx=x +C\)

The third integral, found using the same rule as the first integral, is equal to

\(\displaystyle \int x^{\frac{1}2{}} dx=\frac{2x^{\frac{3}{2}}}{3}+C\)

Combining the results - and adding all the constants of integration to make a single C - we get

\(\displaystyle \frac{(x-5)^3}{3}-20x-\frac{2x^{\frac{3}{2}}}{3}+C\).

Find the area of the region bound by the lines \(\displaystyle x=5\), \(\displaystyle x=10\), the x-axis, and the function v(x)

\(\displaystyle v(x)=14x^6+5x^4-3x^2+4\)

First, we need to set up our integral.

It should look like this:

\(\displaystyle \int_{5}^{10}14x^6+5x^4-3x^2+4dx\)

Next, recall the rule for integrating polynomials: Increase each exponent by 1 and divide by that number:

\(\displaystyle \int_{5}^{10}14x^6+5x^4-3x^2+4dx=\frac{14x^7}{7}+\frac{5x^5}{5}-\frac{3x^3}{3}+4x+c\mid_5^{10}\)

Simplify this to get

\(\displaystyle 2x^7+x^5-x^3+4x+c\mid_5^{10}\)

Next, evaluate our integral by finding the difference between V(10) and V(5)

\(\displaystyle 2x^7+x^5-x^3+4x+c\mid_5^{10}\)

\(\displaystyle (2*{10}^7+{10}^5-{10}^3+4*{10}+c)-(2*{5}^7+{5}^5-{5}^3+4*{5}+c)=20099040-159250=19939770\)

So our answer is 19939770