What is #f(x) = int e^(5x-1)+x dx# if #f(2) = 3 #?

Question

What is #f(x) = int e^(5x-1)+x    dx# if #f(2) = 3  #?

Emma Aldridge · Accepted Answer

#f(x)=1/5(e^(5x-1)+5-e^9)#

#f(x)=int(e^(5x-1)+x)dx" "f(2)=3#

now #d/(dx)(e^(5x-1))=5e^(5x-1)#

using the chain rule. (This is left as an exercise for the reader to show)

by inspection therefore we have

#f(x)=int(e^(5x-1)+x)dx=1/5e^(5x-1)+1/2x^2+C#

but #f(2)=3#

#:.3=1/5e^((5xx2-1))+1/2xx2^2+C#

#3=1/5e^9+2+C#

#15=e^9+10+5C#

#:.C=(5-e^9)/5#

#f(x)=1/5e^(5x-1)+(5-e^9)/5#

#f(x)=1/5(e^(5x-1)+5-e^9)#

Christopher Allers · Answer

undefined To find $ f(x) = \int e^{5x - 1} + x \, dx $ given that $ f(2) = 3 $, we first need to find the antiderivative of the function $ e^{5x - 1} + x $, and then use this antiderivative to evaluate $ f(x) $ at $ x = 2 $.

The antiderivative of $ e^{5x - 1} $ is $ \frac{1}{5} e^{5x - 1} $ and the antiderivative of $ x $ is $ \frac{1}{2}x^2 $. Therefore, the antiderivative of $ e^{5x - 1} + x $ is $ \frac{1}{5} e^{5x - 1} + \frac{1}{2}x^2 $.

Now, to find $ f(x) $, we integrate this antiderivative:

$$ f(x) = \frac{1}{5} \int e^{5x - 1} dx + \frac{1}{2} \int x \, dx $$
$$ f(x) = \frac{1}{5} \left( \frac{1}{5} e^{5x - 1} \right) + \frac{1}{2} \left( \frac{1}{2} x^2 \right) + C $$
$$ f(x) = \frac{1}{25} e^{5x - 1} + \frac{1}{4} x^2 + C $$

Given that $ f(2) = 3 $, we can find the value of $ C $:

$$ 3 = \frac{1}{25} e^{5(2) - 1} + \frac{1}{4} (2)^2 + C $$
$$ 3 = \frac{1}{25} e^9 + \frac{1}{4} \times 4 + C $$
$$ 3 = \frac{1}{25} e^9 + 1 + C $$

Therefore, $ C = 3 - \frac{1}{25} e^9 - 1 $.

Now we substitute this value of $ C $ back into the equation for $ f(x) $:

$$ f(x) = \frac{1}{25} e^{5x - 1} + \frac{1}{4} x^2 + 3 - \frac{1}{25} e^9 $$

So, $ f(x) = \frac{1}{25} e^{5x - 1} + \frac{1}{4} x^2 + 3 - \frac{1}{25} e^9 $.

Scarlett Allison · Answer

undefined To find $ f(x) = \int e^{5x-1} + x \, dx $ given that $ f(2) = 3 $, we first need to find the indefinite integral of $ e^{5x-1} + x $ with respect to $ x $. Then, we can evaluate the function at $ x = 2 $ and solve for the constant of integration.

Given:
$$ f(x) = \int e^{5x-1} + x \, dx $$

First, let's find the indefinite integral:
$$ \int e^{5x-1} + x \, dx $$

$$ = \int e^{5x-1} \, dx + \int x \, dx $$

$$ = \frac{1}{5} e^{5x-1} + \frac{x^2}{2} + C $$

Now, we evaluate the function at $ x = 2 $ using the given information that $ f(2) = 3 $:
$$ f(2) = \frac{1}{5} e^{5(2)-1} + \frac{2^2}{2} + C $$

$$ 3 = \frac{1}{5} e^{9} + 2 + C $$

$$ 3 - 2 = \frac{1}{5} e^{9} + C $$

$$ 1 = \frac{1}{5} e^{9} + C $$

$$ C = 1 - \frac{1}{5} e^{9} $$

Now, we have found the constant of integration ($ C $), so we can write the function $ f(x) $ as:
$$ f(x) = \frac{1}{5} e^{5x-1} + \frac{x^2}{2} + \left(1 - \frac{1}{5} e^{9}\right) $$

Therefore, $ f(x) = \frac{1}{5} e^{5x-1} + \frac{x^2}{2} + 1 - \frac{1}{5} e^{9} $.