Add the numbers from 1 to 100 by following the

Ответ:

sairaanwar67

03.01.2022

Summaation of a sequence

or see a smaller scale
from 1 to 10
notice
1+10=11
2+9=11
etc
5+6=11
10/2=5
there are 5 pairs
5*11=55

so

1 to 100
1+100=101
2+99=101
...
50+51=101

100/2=50
there ar 50 pairs
50 times 101=5050

the sum is 5050

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Ответ:

ZillaKami

29.12.2020

Mathematics

Answer is $\displaystyle\int \frac{\sec^2\left(\frac{1}{x^9}\right)}{ x^{10}} \, dx = -\frac{1}{9}\tan \left(\frac{1}{x^9}\right)+c$ ======i assume this is what this is what question asks for as it makes the substitution work: $\displaystyle\int \frac{\sec^2\left(\frac{1}{x^9}\right)}{ x^{10}} \, dx$ if $u = \dfrac{1}{x^9} = x^{-9}$ then by simple differentiation, we have $\begin{aligned} du & = -9x^{-10} dx \\ du & = -\frac{9}{x^{10}} dx\\ -\frac{x^{10}}{9} du & = dx \\ dx & = -\frac{x^{10}}{9} du \end{aligned}$ note that $\int \sec^2 u \; du = \tan u + c$ therefore, performing these substitutions: $\begin{aligned} \int \frac{\sec^2\left(\frac{1}{x^9}\right)}{ x^{10}} \, dx & = \int \frac{\sec^2\left(u\right)}{ x^{10}} \cdot \left(-\frac{x^{10}}{9}\right) \; du \\ & = \int \sec^2\left(u\right) \cdot \left(-\frac{1}{9}\right)\; du \\ & = -\frac{1}{9}\int \sec^2\left(u\right) \; du \\ & = -\frac{1}{9}\tan u + c \\ & = -\frac{1}{9}\tan \left(\frac{1}{x^9} \right) + c \end{aligned}$

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Add the numbers from 1 to 100 by following the same steps you used in 1.

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