A)
$$\int_0^{ln(3)} \frac{e^x}{e^x +2}dx$$
$$u=e^x +2$$ $$du = e^x$$
$$\int_0^{ln(x)} u^{-1}du$$
when x = ln(3), u = 5
when x = 0, u = 3
$$ln(e^x + 2)]_3^5 = ln|e^5 +2| - ln|e^3+2| = 1.9184629453$$
without reevaluating limits I get a different answer
$$ln(e^x + 2)]_0^{ln(3)} = ln|e^{ln(3)}+2| - ln|e^0 + 2| = 0.510825623766$$
when do I know to reevaluate limits? If I get a different answer for both, which one is correct?
B)
$$\int_0^{\frac{-\sqrt{3}}{5}} \frac{arctan(5x)}{1+25x^2}dx$$
$$u=arctan(5x)$$ $$du=\frac{5}{1+25x^2}$$
$$\frac{1}{5}\int_0^{\frac{-\sqrt{3}}{5}}udu$$
I reevaluated the limits in the integral, because of its form 'udu'
I'm just confused when to reevaluate the limits, because I know its not always necessary. But doing so shouldn't you end up with the same answer if you didn't? what form does the integral have to take for me to know I have to, or its ok to reevaluate the limits. An answer without too much terminology would be helpful, thanks
what if it takes the form of:
$$\int_3^{ln(9)} e^u du$$
for example?
You have two choices. You can reevaluate the limits, or you can express the integral in terms of the original variable and use the original limits. Taking your second example, maybe it helps to show the limits like this: $$\int_{x=0}^{x=\frac{-\sqrt{3}}{5}} \frac{\arctan(5x)}{1+25x^2}dx$$ $$u=\arctan(5x)$$ $$du=\frac{5}{1+25x^2}$$
$$\frac{1}{5}\int_{x=0}^{x=\frac{-\sqrt{3}}{5}}udu=\frac 15\cdot\left.\frac {u^2}2\right|_{x=0}^{x=\frac{-\sqrt{3}}{5}}$$ Now it is clear what our choices are. One is to reevaluate the limits: $$\frac 15\left.\frac {u^2}2\right|_{x=0}^{x=\frac{-\sqrt{3}}{5}}=\frac 15\cdot\left.\frac {u^2}2\right|_{u=0}^{u=\frac{-\pi}{3}}=\frac{\pi^2}{90}$$ and the other is to back substitute: $$\frac 15\left.\frac {u^2}2\right|_{x=0}^{x=\frac{-\sqrt{3}}{5}}=\left.\frac 1{10}(\arctan(5x))^2\right|_{x=0}^{x=\frac{-\sqrt{3}}{5}}=\frac{\pi^2}{90}$$ Depending on the problem, one may be easier than the other.