In order to find the area underneath the curve to the x-axis on the interval \(\displaystyle [a,b]\) we must solve the integral

\(\displaystyle \int_{a}^{b} \left | f(x)\right |\, dx\).

Because the function is always positive on the interval \(\displaystyle [0,2]\) we solve the integral

\(\displaystyle \int_{0}^{15} x^4 \, dx\).

When taking the integral we apply the inverse power rule which states

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

As such

\(\displaystyle \int_{0}^{2} x^4 \, dx = \frac{x^5}{5}|_0^{2}\).

And by the corollary of the first Fundamental Theorem of Calculus

\(\displaystyle \frac{x^5}{5}|_0^{2}=\frac{2^5}{5}-0=\frac{32}{5}\).

As such the area is

\(\displaystyle \frac{32}{5}\) square units.

In order to find the area underneath the curve to the x-axis on the interval \(\displaystyle [a,b]\) we must solve the integral

\(\displaystyle \int_{a}^{b} \left | f(x)\right |\, dx\).

Because the function is always positive on the interval \(\displaystyle [0,1]\) we solve the integral

\(\displaystyle \int_{0}^{1} e^x \, dx\).

Because the antiderivative of the exponential function is the exponential function itself, we obtain 

\(\displaystyle \int_{0}^{1} e^x \, dx = e^x|_0^1\).

And by the corollary of the first Fundamental Theorem of Calculus

\(\displaystyle e^x|_0^1 = e^1-e^0=e-1\).

As such the area is

\(\displaystyle e-1\) square units.

Evaluate the following integral to find the area of the region bound by the function and the given limits.

\(\displaystyle \int_{0}^{5}x^4-6x^2+8x+6dx\)

Begin by recalling the integration rule for polynomials. All we need to do to integrate a term of a polynomial is raise the exponent by 1 and divide by the new number.

So, we go from this

\(\displaystyle \int_{0}^{5}x^4-6x^2+8x+6dx\)

To this

\(\displaystyle \frac{x^5}{5}-\frac{6x^3}{3}+\frac{8x^2}{2}+6x+c|_0^5\)

We now have successfully integrate our function, but we still must evaluate it with the given limits. To do so, simply plug in our lower and upper limits and find the difference between them. Note, this is step is simplified by the fact that our lower limit is 0, because plugging zero into our function will just yield "c"

So...

\(\displaystyle \frac{5^5}{5}-\frac{6(5)^3}{3}+\frac{8(5)^2}{2}+6(5)+c-c\)

\(\displaystyle 5^4-2*5^3+4*5^2+30=505\)

Making our answer 

\(\displaystyle 505 units^2\)

Find the area bound by the x and y axes, the function g(x), and the line \(\displaystyle x=6\)

\(\displaystyle g(x)=8x^3+6x^2-5x+3\)

Because we are asked to find area, we want to set up an integral. Be sure to include the correct limits of integration:

\(\displaystyle \int_{0}^{5}g(x)dx=\int_{0}^{5}8x^3+6x^2-5x+3dx\)

\(\displaystyle \int_{0}^{5}8x^3+6x^2-5x+3dx=2x^4+2x^3-\frac{5x^2}{2}+3x+c\mid_0^5\)

\(\displaystyle 2x^4+2x^3-\frac{5x^2}{2}+3x+c\mid_0^5=2(5)^4+2(5)^3-\frac{5(5)^2}{2}+3(5)+c-c\)

\(\displaystyle 2(5)^4+2(5)^3-\frac{5(5)^2}{2}+3(5)=1250+250-\frac{125}{2}+15=1452.5\)

So our final answer is:

\(\displaystyle 1452.5units^2\)

Find the intersection points created by the bounded lines.  The intersection of \(\displaystyle x=1\), \(\displaystyle y=x\),  and \(\displaystyle y=1\) is \(\displaystyle (1,1)\).    The intersection of lines \(\displaystyle y=1\) and \(\displaystyle x=5\) is \(\displaystyle (5,1)\).

Since the area does not include the region below \(\displaystyle y=1\), we must subtract the areas of the top curve with the area of the bottom curve.  The top curve is \(\displaystyle y=x\) and the bottom curve is \(\displaystyle y=1\).

Integrate the function \(\displaystyle y=x\) with the respect to \(\displaystyle x\) from \(\displaystyle 1\) to \(\displaystyle 5\).

\(\displaystyle \int_{1}^{5}x\:dx = \frac{x^2}{2} |_{1}^{5} = \frac{5^2}{2}-\frac{1^2}{2} = \frac{25}{2}-\frac{1}{2}= 12\)

Integrate the function \(\displaystyle y=1\) with the respect to \(\displaystyle x\) from \(\displaystyle 1\) to \(\displaystyle 5\).

\(\displaystyle \int_{1}^{5} 1\:dx = x|_{1}^{5} = 5-1 =4\)

Subtract both areas to get the area of the bounded region.

\(\displaystyle A=12-4 = 8\)

To solve, simply integrate the function from \(\displaystyle -2\) to \(\displaystyle 2\). Thus,

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

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

\(\displaystyle =(14)-(-6)=20\)

Since \(\displaystyle \sqrt {9-x^2}\) is only defined when \(\displaystyle 9-x^2 \geq 0 \Rightarrow -3 \leq x \leq 3.\)\(\displaystyle \int_{-\infty}^{\infty} \sqrt{9-x^2} \ dx = \int_{-3}^{3} \sqrt{9-x^2} \ dx.\) Also, the graph of the function \(\displaystyle y=\sqrt {9-x^2}\) is the upper half of the circle with center as \(\displaystyle (0, 0)\) and radius \(\displaystyle 3\). By definition of integral, we can find the value of \(\displaystyle \int_{-3}^{3} \sqrt{9-x^2} \ dx\) is the same as the area of the semi circle. Therefore, it is equal to \(\displaystyle \frac{1}{2} \pi r^2 = \frac{1}{2} \pi \cdot 3^2= \frac{9 \pi}{2}.\)

To solve, simply integrate from \(\displaystyle x=0\) to \(\displaystyle x=3\). Thus,

\(\displaystyle \int_{0}^{3}2x-1\ dx=x^2-x\Big |_{0}^{3}=(3^2-3)-(0^2-0)=9-3=6\)

To find the area between two curves, we first need to determine on what interval we will be taking their integrals. This interval has endpoints where the two functions intersect, so we need to find the intersection points.

To find intersection points of two curves, set the two functions equal (since the \(\displaystyle \small y\)-values are the same at the intersection points) and solve for the \(\displaystyle \small x\)-value(s).

Here:

\(\displaystyle \small x^2 + 2x + 1 = -3x +1\)

\(\displaystyle \small x^2 + 5x = 0\)

\(\displaystyle \small x(x + 5) = 0\)

So \(\displaystyle \small x=0\) or \(\displaystyle \small x=-5\), and we integrate over the interval \(\displaystyle \small [-5, 0]\).

We also need to know which curve is above the other. Find this by graphing or by testing \(\displaystyle \small y\)-values in the interval of interest. In this case, the linear function is the uppermost.

Then take the integral of the upper function (the linear here), and subtract the integral of the lower function (the quadratic here), so that we subtract out the area we do not want to count.

To find the area under the curve, we need to take the definite integral. But in order to do that, we first need to determine the limits of integration.

In this case, since the graph of the function is a parabola opening downward, we need to find the places where it intersects the \(\displaystyle x\)-axis -- the zeros of the function.

To find zeros, set the function equal to 0 and solve for \(\displaystyle x\):

\(\displaystyle f(x) = -x^2 + 4x - 3 = 0\)  //  divide by -1 to make factoring easier

\(\displaystyle x^2 - 4x + 3 = 0\)  //  factor

\(\displaystyle (x - 3)(x - 1) = 0\)  //  solve

\(\displaystyle x=3\) or \(\displaystyle x=1\)

So we want to take the integral from \(\displaystyle x=1\) to \(\displaystyle 3\) of \(\displaystyle f(x)\).

The antiderivative is \(\displaystyle \small F(x) = -\frac{1}{3}x^3 + 2x^2 - 3x + C\). (We omit \(\displaystyle C\) when finding the definite integral.)

We want \(\displaystyle F(3) - F(1)\), or

\(\displaystyle \small \left[-\frac{1}{3}(3)^3 + 2(3)^2 - 3(3)\right] - \left[-\frac{1}{3}(1)^3 + 2(1)^2 - 3(1)\right]\)

\(\displaystyle \small = [-9 + 18 - 9] - \left[-\frac{1}{3} +2 -3\right]\)

\(\displaystyle \small = [0] - \left[-\frac{4}{3}\right]\)

\(\displaystyle =\frac{4}{3}\)