definite integrals
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Harry_the_cat
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Note that \(\displaystyle \frac{d}{dt}ln(1+0.02t) =\frac{0.02}{1+0.02t}\). That should help. Can you proceed from there?
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Dr.Peterson
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The method that has been suggested is a good one, based on seeing a likely form for the antiderivative, and making adjustments as needed. You should try doing what you were asked to do, and show how far you get.
But if you need to follow a method you have been taught, rather than try something and seeing what happens, do you know how to use substitution (sometimes called "u-substitution")? Let u = 1 + 0.02t, and follow the procedure as far as you can. We can help you along, but you need to do some work in order for help to succeed.
But if you need to follow a method you have been taught, rather than try something and seeing what happens, do you know how to use substitution (sometimes called "u-substitution")? Let u = 1 + 0.02t, and follow the procedure as far as you can. We can help you along, but you need to do some work in order for help to succeed.