Question

Integrate from -5 to 5 |x+2| dx

please give me the steps also ​

Answer 1

Answer:

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Steps in evaluating the integral of complementary error function?

integration special-functions

Could you please check the below and show me any errors?

∫∞xerfc (t) dt =∫∞x[2π−−√∫∞te−u2du] dt

If I let dv=dt and u equal the term inside the bracket, and do integration by parts,

∫u dv =uv−∫v du

v=t and du becomes

−2π−−√e−t2

This was obtained from using the Leibniz rule below,

ddt[∫baf(u)du] =∫baddtf(u)du+fdbdt−fdadt

Then,

ddt[2π−−√∫∞te−u2du] =2π−−√[∫∞tddt(e−u2)du+e−∞2∗0−e−t2∗1]=2π−−√[0 + 0 −e−t2]

Is the first and second term going to zero correct? The upper limit b=infinity, and is db/dt=0 in the second term correct?

The integral becomes

[ t 2π−−√∫∞te−u2du ]∞x+∫∞xt[2π−−√e−t2] dt=

[ t 2π−−√∫∞te−u2du ]∞x−[1π−−√e−t2]∞x=

[0− x 2π−−√∫∞xe−u2du ]−[0−1π−−√e−x2]=

Answer 2

GIVEN:-

$\sf \int \limits_{ - 5}^{5} |x + 2| dx$

SOLUTION :-

$\sf \int \limits_{ - 5}^{5} |x + 2| dx$

• Here first we have to find the limits

$\sf \: |x + 2| = \begin{cases} \sf - (x + 3)& \sf{if\: x < - 3} \\ \sf \: \: \: \: (x + 3) & \sf{if\: x > - 3 }\end{cases}$

$\sf \int \limits_{ - 5}^{ 5} |x + 2| dx =$

$\sf \: \int \limits_{ - 5}^{ - 3} - ( x + 2)dx + \int \limits_{ - 3}^{5} (x + 2) dx$

$= \sf \: - \int \limits_{ - 5}^{ - 3} ( x + 2)dx + \int \limits_{ - 3}^{5} (x + 2) dx$

$= \sf - \bigg( \frac{ {x}^{2} }{2} + 2x \bigg)^{ - 3} _{ - 5} + \bigg( \frac{ {x}^{2} }{2} + 2x \bigg)^{ 5} _{ - 3}$

$= \sf - \bigg( \frac{ {( - 3)}^{2} }{2} + 2( - 3) - \frac{( - 5) {}^{2} }{2} - 2( - 5) \bigg)+ \\ \\ \sf\bigg( \frac{ {5}^{2} }{2} + 2(5) - \frac{( - 3) {}^{2} }{2} - 2( - 3) \bigg)$

$= \sf - \bigg( \frac{ 9}{2} - \frac{ 2 5 }{2} - 6 + 10 \bigg)+ \sf\bigg( \frac{ 25 }{2}- \frac{9 }{2} + 10 + 6 \bigg)$

$= \sf \bigg( - \frac{ 9}{2} - \frac{9 }{2} + \frac{ 2 5 }{2} + \frac{ 2 5 }{2} + 6 - 10 + 10 + 6 \bigg)$

$= \sf \: - 9 + 25 + 12$

$= \sf \: 28$

Ans.....