Saladsamurai
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Definite Integrals
I think I got these, but I left my text at work, so I was hoping someone could confirm that these answers are correct?
a.)
\int_{-1}^{1}\frac{dx}{1+x^2}
=\tan^{-1}(1)-\tan^{-1}(-1) =-\frac{\pi}{2}
and
b.)
\int_0^{\ln5}e^x(3-4e^x)
=\int_0^{\ln5}[3e^x-4e^{2x}]dx
=3e^x-2e^{2x}]_0^{\ln5}=49
Thanks,
Casey
I think I got these, but I left my text at work, so I was hoping someone could confirm that these answers are correct?
a.)
\int_{-1}^{1}\frac{dx}{1+x^2}
=\tan^{-1}(1)-\tan^{-1}(-1) =-\frac{\pi}{2}
and
b.)
\int_0^{\ln5}e^x(3-4e^x)
=\int_0^{\ln5}[3e^x-4e^{2x}]dx
=3e^x-2e^{2x}]_0^{\ln5}=49
Thanks,
Casey
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