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2 years ago

I need HELP PLZZZZ THIS IS DUE AT 7:00

Mathematics

2 answers:

valkas [14]2 years ago

7 0

Answer:

27.5

Step-by-step explanation:

1.5(5)+20=27.5

:) hope its right :)

Step2247 [10]2 years ago

7 0

Answer:the answer is 2/5

Step-by-step explanation:

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$\large\begin{array}{l} \mathsf{\displaystyle\int\!3^{2x-1}\,dx}\\\\ =\mathsf{\displaystyle\int\!\frac{1}{2}\cdot 2\cdot 3^{2x-1}\,dx}\\\\ =\mathsf{\displaystyle\frac{1}{2}\int\!3^{2x-1}\cdot 2\,dx\qquad(i)} \end{array}$

$\large\begin{array}{l} \textsf{Substitute}\\\\ \mathsf{2x-1=u\quad\Rightarrow\quad 2\,dx=du}\\\\\\ \textsf{so (i) becomes}\\\\ =\mathsf{\displaystyle\frac{1}{2}\int\!3^u\,du}\\\\ =\mathsf{\dfrac{1}{2}\cdot \dfrac{1}{\ell n\,3}\,3^u+C}\\\\ =\mathsf{\dfrac{1}{2\,\ell n\,3}\,3^{2x-1}+C} \end{array}$

$\large\begin{array}{l} \boxed{\begin{array}{c}\mathsf{\displaystyle\int\!3^{2x-1}\,dx=\frac{1}{2\,\ell n\,3}\,3^{2x-1}+C} \end{array}}\qquad\checkmark \end{array}$

If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2154165

$\large\textsf{I hope it helps. :-)}$


Tags: integrate indefinite integral substitution exponential base logarithm log ln composite integral calculus

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