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Personal u-subs

mark

(10 pts)

In this problem, you’re going to compute your own personal definite integral using u-substitution with a change in the bounds of integration. To get your integral, choose your name from the following list:

After you have your integral, respond to this question by hitting the Reply button below. Be sure to use LaTeX to:

Type out your personal integral (the statement of your problem includes the LaTeX code for your integral).
Indicate the substitution you intend to use - including the du portion, as well as the u
Type out the complete solution to the problem and be sure to change the bounds of integration in the process.

Finally, you should include a graph that illustrates the integral.

audrey

My integral is:

\int_0^4 2 x \cos \left(x^2\right) \, dx.

I’ll let u= x^2 so that du = 2x \, dx. Thus,

\int_0^4 2 x \cos \left(x^2\right) \, dx = \int_0^{16} \cos(u) \, du = \sin(u)|_0^{16} = \sin(16).

The graph looks like so:

$$
\int_0^4 2 x \cos \left(x^2\right) \, dx = \int_0^{16} \cos(u) \, du = \sin(u)|_0^{16} = \sin(16).
$$

(1/16) = \frac{1}{16}

$$
  (1/16) = \frac{1}{16}
$$

\frac{1}{8} \int x^2 \, dx = (1/8) \int x^2 \, dx.

fcarrill

My integral is:

\int_0^3 2 x \sin \left(x^2\right) \, dx

I’ll let u=x^2 so that du = 2x \,dx. Thus,

\int_0^3 2 x \sin \left(x^2\right) \, dx = \int_0^{9} \ sin(u) du=-cos(u)|_0^9 =-cos(9) +cos(0)

The graph looks like so:

mearing

My integral is:

\int_1^4 2 e^{x^2} x \, dx

I’ll let u=x^2 so that du=2xdx

\int_1^4 2 e^{x^2} x \, dx = \int_1^{16} e^{u}du= e^u|_1^{16}= e^{16}-e^1

The graph looks like

afernan2

My intergral is:

\int_1^3 e^{3 x} \, dx

u=3x ,du=3dx ,du/3=dx

Thus the intergral now becomes

\int_1^3 e^{u} \, dx = \int_1^3 e^{3x}/3 \, dx \, = \, \frac{e^{3x}}{3}|_{e/3}^{{e^9}/3}=2694.333

I believe that is in the correct format.

knguyen3

My personal integral is \int_{-1}^3 \cos ^3(x) \sin (x) \, dx
I’ll let u = cos(x) so that du = -sin(x) dx. Thus,

\int_{-1}^3 \cos ^3(x) \sin (x) \, dx = \int_{sin(1)}^{-sin(3)}\ -u^3 \, du = -\frac{u^4}{4}|_{sin(1)}^{-sin(3)} = -\frac{sin(3)^4}{4} + \frac{sin(1)^4}{4} = \frac{sin(1)^4-sin(3)^4}{4}

The graph is:

[]

myost

integral:

\int_1^3 3 e^{x^3} x^2 \, dx

let u=x^3 and du=3x^2dx. thus,

\int_1^3 3 e^{x^3} x^2 \, dx = \int_1^{27} e^{u} \, du = e^u |_1^{27} = e^{27}-e^1

nhaley

My integral is:

\int_0^4 3 x^2 \cos \left(x^3\right) \, dx

I’ll let u=x^3 so the du=3x^2dx. Thus,

\int_0^4 3 x^2 \cos \left(x^3\right) \, dx = \int_0^{64} \cos(u)du = \sin(u)|_0^{64}=\sin(64).

The graph looks like:

gwebb

My integral is:

\int_1^2 \cos (x) \sin ^3(x) \, dx

I’ll let

u = sin(x)

so that

du = cos(x) dx

Thus,

\int_1^2 \cos (x) \sin ^3(x) \, dx = \int_{\sin (1)}^{\sin (2)} u ^3(x) \, du = \frac{u^4}{4} |_{\sin (1)}^{\sin (2)} = \frac{sin(2)^4}{4} + \frac{sin(1)}{4} = \frac{sin(1)^4- sin(2)^4}{4}

ccase2

My integral is:

\int_1^3 \cos (x) \sin ^2(x) \, dx

I’ll let u=sin(x) so that du=cos(x)\,dx. Thus,

\int_1^3 \cos(x) \sin^2(x) \, dx = \int_{sin(1)}^{sin(3)} \ u^2 \, du = \frac{u^3}{3}|_{sin(1)}^{sin(3)} = \frac {sin^3(3) - sin^3(1)}{3}

The graph looks like so:

tbrincke

My integral is:

\int_0^4 \cos ^2(x) \sin (x) \, dx

I’ll let u = cos(x) so that du = -sin(x) dx Thus,

\int_0^4 \cos ^2(x) \sin (x) \,dx = \int_{cos(0)}^{cos(4)} -u^2du= \frac{-u^3}{3} |_{cos(0)}^{cos(4)} = \frac{-cos^3(4)}{3} + \frac{1}{3}

wnovak

My integral is

\int_{-1}^3 3 x^2 \sin \left(x^3\right) \, dx

Let u=x^{3} , du=3xdx

\int_{-1}^3 3 x^2 \sin \left(x^3\right) \, dx =\int_{-1}^{27} \sin (u)du = - \cos u |_{-1}^{27}=-\cos(27)+\cos(1)

agooch

My integral is:
\int_0^4 \cos ^4(x) \sin (x) \, dx
I’ll let u= cos(x) so that du= -sin(x)dx . Thus,

\int_0^4 \cos ^4(x) \sin (x) \, dx = -\int ^{\cos(4)}_0u^4du = -\frac{u^5}{5} = -\frac{cos^5(x)}{5} |_0^{cos(4)} = -\dfrac{\cos^5\left(\cos\left(4\right)\right)-1}{5}

The graph of it is shown below:

chowell1

My integral is:

\int_{-1}^2 4 x^3 \sin \left(x^4\right) \, dx

I’ll let u = x^4 so that du =4x^3 \, dx. Thus,

\int_{-1}^2 4 x^3 \sin \left(x^4\right) \, dx = \int_{1}^{16} \sin \left(u\right) \, dx \, = -\cos \left(u\right)|_1^{16}= -cos(16) + cos(1) = 1.5 \, .

The graph looks like so:

cbrowni1

My integral is…

\int_1^3 4 x^3 \cos \left(x^4\right) \, dx

u={x^4}

du=4{x^3}dx

Then…

\int_1^{81} \cos \left(u\right) \, du= sin(u)|_1^{81}=sin(81)-sin(1)

The graph of this is shown below…

gbelk

My integral is:

\int_1^2 \cos (x) \sin ^3(x) \, dx

I’ll let u = sin(x) so that du = -cos(x) dx . Thus

\int_1^2 \cos (x) \sin ^3(x) \, dx = \int_ {\sin(1)} ^ {\sin(2)} u^3 du = - u^4/4 = -\sin ^4(x)/4|_{\sin(1)} ^ {\sin(2)} = -\sin^4(\sin(2))/4 + \sin^4(\sin(1))/4

rstahles

My personal integral is:

\int_0^3 4 e^{x^4} x^3 \, dx.

I’ll let u = x^4, so that du = 4x^3dx. Thus,

\int_0^3 4 e^{x^4} x^3 \, dx = \int_0^{81} e^{u} \, du = e^u|_0^{81} = e^{81}.

That equals about 1.51 x 10^{35}. The graph of this looks like:

rtaylor4

My integral is:

\int_0^3 e^{5 x} \, dx

I’ll let u = 5x so that du = 5dx.

\int_0^3 e^{5 x} \, dx = (1/5)\int_0^{15} e^{u} du \ = (1/5) [e^{u} ] |_0^{15} = (e^{15} -1)/5

jlajcin

My integral is:

\int_0^4 \cos (x) \sin ^4(x) \, dx

I’ll let u=\sin(x) so that du=cos(x)dx
Thus

\int_0^4 \cos (x) \sin ^4(x) \, dx = \int_0^{\sin(4)} u^4du =\int_0^{\sin(4)}(1/5)u^5du|_0^{sin(4)}=(1/5)sin^5(4)

mark