First, set up the integral expression:

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

Now, integrate. Remember to raise the exponent by 1 and then also put that result on the denominator:

\(\displaystyle \frac{2x^4}{4}-\frac{2x^2}{2}+x=\frac{x^4}{2}-x^2+x\)

Evaluate at 2 and then 1. Subtract the results:

\(\displaystyle (8-4+2)-(\frac{1}{2}-1+1)=6-\frac{1}{2}=5.5\)

First, set up the integral expression:

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

Now, integrate. Remember to raise the exponent by 1 and then also put that result on the denominator:

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

Evaluate at 5 and then 4. Subtract the results:

\(\displaystyle (\frac{125}{3}-\frac{125}{2}+20)-(\frac{64}{3}-\frac{80}{2}+16)\)

Simplify to get your answer of: \(\displaystyle \frac{11}{6}\)

First, set up the integral expression:

\(\displaystyle \int_{4}^{5}x^2+4x+4dx\)

Next, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

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

Now, evaluate at 5 and then 4. Subtract the results:

\(\displaystyle (\frac{125}{3}+50+20)-(\frac{64}{3}+32+16)=22-\frac{61}{3}=\frac{5}{3}\)

First, write out the integral expression:

\(\displaystyle \int_{0}^{1}x^2-7x+9dx\)

Next, integrate. Remember to add one to the exponent and also put that result on the denominator

\(\displaystyle \frac{x^3}{3}-\frac{7x^2}{2}+9x\)

Next, evaluate at 1 and then 0. Subtract the results:

\(\displaystyle (\frac{1}{3}-\frac{7}{2}+9)-(0)=\frac{35}{6}\)

First, write out the integral expression for this problem:

\(\displaystyle \int_{3}^{4}x^3-2x^2+1dx\)

Next, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

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

Next, evaluate at 4 and then 3. Subtract the results:

\(\displaystyle (\frac{256}{4}-\frac{128}{3}+4)-(\frac{81}{4}-\frac{54}{3}+3)\)

Simplify to get your answer:

\(\displaystyle \frac{175}{4}-\frac{74}{3}+1=\frac{233}{12}\)

The area under the given curve is found using the following integral:

\(\displaystyle \int_{0}^{2}(e^{2x}+x^2+2)dx=\frac{e^{2*2}}{2}+\frac{2^3}{3}+2*2-\frac{e^0}{2}=\frac{e^4-1}{2}+20/3\)

Find the area of the region bounded by the curves \(\displaystyle \small x = -y^2\) and \(\displaystyle \small y = -x-2\)

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Definition

The area between two curves over the interval \(\displaystyle a\leq y\leq b\)  is defined by the integral: 

\(\displaystyle Area =\int_a^b [f(y)_{upper }-g(y)_{lower}]dy\)

where \(\displaystyle f(y)_{upper}\geq g(y)_{lower}\) for all \(\displaystyle y\)  such that  \(\displaystyle a\leq y\leq b\)

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Looking at the graphs of the curves we were given, we quickly see that integrating with respect to \(\displaystyle y\) would be easiest. Attempting to integrate with respect to \(\displaystyle x\), we would not be able to assert \(\displaystyle f(x)_{upper}\geq g(x)_{lower}\) for all values of \(\displaystyle x\) between the intersection points. In fact, \(\displaystyle \small x = -y^2\) would not even pass the vertical line test to be considered a function. 

First let's find where the curves intersect. The \(\displaystyle y\) coordinate of both points will be our limits of integration. Substitute the equation for the parabola into the equation of the line we were given: 

 \(\displaystyle y=(-y^2)-2\)

 \(\displaystyle y^2 - y-2=0\)

\(\displaystyle (y+1)(y-2)=0\)

The solutions are therefore \(\displaystyle y = \left \{ -1,2\right \}\)

 \(\displaystyle Area =\int_a^b [f(y)_{upper }-g(y)_{lower}]dy\)

\(\displaystyle \small Area = \int_{-1}^2[-y^2-(-y-2)]dy\)

\(\displaystyle \small \small Area = \int_{-1}^2[-y^2+y+2)]dy\)

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

\(\displaystyle \small Area = \left(-\frac{8}{3}+2+4 \right)-\left(\frac{1}{3}+\frac{1}{2}-2 \right )\)

\(\displaystyle \small \small Area = \left(\frac{10}{3} \right)-\left(-\frac{7}{6} \right )\)

\(\displaystyle \small Area = \frac{9}{2}\)

From calculus, we know the volume of an irregular solid can be determined by evaluating the following integral:

\(\displaystyle V=\int_{a}^{b}A(x)dx\)

Where A(x) is an equation for the cross-sectional area of the solid at any point x. We know our bounds for the integral are x=1 and x=4, as given in the problem, so now all we need is to find the expression A(x) for the area of our solid.

From the given bounds, we know our unrotated region is bounded by the x-axis (y=0) at the bottom, and by the line y=x^2-4x+5 at the top. Because we are rotating about the x-axis, we know that the radius of our solid at any point x is just the distance y=x^2-4x+5. Now that we have a function that describes the radius of the solid at any point x, we can plug the function into the formula for the area of a circle to give us an expression for the cross-sectional area of our solid at any point:

\(\displaystyle A(x)=\pi r^2=\pi (x^2-4x+5)^2=\pi (x^4-8x^3+26x^2-40x+25)\)

We now have our equation for the cross-sectional area of the solid, which we can integrate from x=1 to x=4 to find its volume:

\(\displaystyle V=\int_{1}^{4}\pi (x^4-8x^3+26x^2-40x+25)dx\)

\(\displaystyle =\pi [\frac{1}{5}x^5-2x^4+\frac{26}{3}x^3-20x^2+25x]_{1}^{4}=\frac{78\pi }{5}\)

Write the formula for cylindrical shells, where \(\displaystyle r\) is the shell radius and \(\displaystyle h\) is the shell height.

\(\displaystyle V=\int_{a}^{b} 2\pi rh \:dy\)

Determine the shell radius.

\(\displaystyle r=y\)

Determine the shell height.  This is done by subtracting the right curve, \(\displaystyle x=4\), with the left curve, \(\displaystyle x=y^2\).

\(\displaystyle h=4-x=4-y^2\)

Find the intersection of \(\displaystyle x=y^2\) and \(\displaystyle x=4\) to determine the y-bounds of the integral.

\(\displaystyle 4=y^2\)

\(\displaystyle 2=y\)

The bounds will be from 0 to 2.  Substitute all the givens into the formula and evaluate the integral.

\(\displaystyle V=\int_{a}^{b} 2\pi rh \:dy = \int_{0}^{2} 2\pi \cdot y\cdot \left ( 4-y^2)dy\)

\(\displaystyle V= 2\pi \int_{0}^{2} \left ( 4y-y^3\right ) dy\)

\(\displaystyle V=2\pi \left[ 2y^2 - \frac{1}{4}y^4\right]_{0}^{2} = 2\pi \left[2(2)^2 -\frac{1}{4}(2)^4\right] = 2\pi[8-4]= 8\pi\)

Since we are revolving a region of interest around a horizontal line \(\displaystyle {}y = 0\), we need to express the inner and outer radii in terms of x.

Recall the formula:

\(\displaystyle Volume = \pi \int_{b}^{a} [(\textup{outer radius})^{2} - (\textup{inner radius})^{2}]dx\)

The outer radius is \(\displaystyle y = sin(x)\) and the inner radius is \(\displaystyle y = 0\). The x-limits of the region are between \(\displaystyle x=0\) and \(\displaystyle x=\pi\). So the volume set-up is:

\(\displaystyle V = \pi \int_{0}^{\pi} sin^{2}x\)

Using trigonometric identities, we know that:    

\(\displaystyle sin^{2}x = \frac{1}{2} - \frac{1}{2}cos 2x\)

Hence:

\(\displaystyle V = \pi \int_{0}^{\pi} sin^2x dx =\pi \int_{0}^{\pi} \left(\frac{1}{2} - \frac{1}{2}cos2x\right) dx\)

\(\displaystyle = \pi \left[\frac{1}{2}x - \frac{1}{4}sin2x\right]_{0}^{\pi} = \frac{1}{2}\pi^2\)