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How do I evaluate #int_0^5(2 e^x + 5cos(x)) dx#?

Answer 1

Emma Aldridge

Hello !

Answer. 2e^5 + 5sin(5) - 2.

Explanation Take antiderivative of #x\mapsto e^x# and #x\mapsto cos(x)# and use the linearity of integral :

#\int_0^5 (2e^x + 5cos(x)) dx = [2e^x + 5sin(x)]_0^5#

#\int_0^5 (2e^x + 5cos(x)) dx = (2e^5 + 5sin(5)) - (2e^0 + 5sin(0))#

#\int_0^5 (2e^x + 5cos(x)) dx = 2e^5 + 5sin(5) - 2#.

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Answer 2

Adam Adkins

To evaluate the integral (\int_{0}^{5} (2e^{x} + 5\cos(x)) , dx), you can integrate each term separately using the properties of integrals. The integral of (e^{x}) is (e^{x}), and the integral of (\cos(x)) is (\sin(x)). Then, evaluate each term at the upper limit (5) and subtract the value of each term evaluated at the lower limit (0). So, the integral evaluates to:

[ \begin{align*} &\int_{0}^{5} (2e^{x} + 5\cos(x)) , dx \ &= \left[2e^{x} + 5\sin(x)\right]_{0}^{5} \ &= (2e^{5} + 5\sin(5)) - (2e^{0} + 5\sin(0)) \ &= 2e^{5} + 5\sin(5) - 2 - 0 \ &= 2e^{5} + 5\sin(5) - 2 \end{align*} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.