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Evaluating a definite integral of an exponential function via FTC Part 2

Evaluate the definite integral ∫−11e2x dx\int_{-1}^{1} e^{2x}\, dx using the antiderivative F(x)=12e2xF(x) = \frac{1}{2}e^{2x} and the Fundamental Theorem of Calculus, Part 2.

Hint

Use the fact that ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a). Here F(x)=12e2xF(x) = \frac{1}{2}e^{2x}, so plug in the upper limit x=1x=1 and the lower limit x=−1x=-1, then subtract.

Step-by-step solution

By FTC Part 2, since F(x)=12e2xF(x) = \frac{1}{2}e^{2x} is an antiderivative of e2xe^{2x}, we have ∫−11e2x dx=F(1)−F(−1)=12e2x∣−11.\int_{-1}^{1} e^{2x}\,dx = F(1) - F(-1) = \left.\frac{1}{2}e^{2x}\right|_{-1}^{1}. This means we evaluate FF at the upper limit x=1x=1 and subtract its value at the lower limit x=−1x=-1.

Substituting the limits gives F(1)=12e2F(1) = \frac{1}{2}e^{2} and F(−1)=12e−2F(-1) = \frac{1}{2}e^{-2}. Therefore ∫−11e2x dx=12e2−12e−2=12(e2−e−2),\int_{-1}^{1} e^{2x}\,dx = \frac{1}{2}e^{2} - \frac{1}{2}e^{-2} = \frac{1}{2}\left(e^{2} - e^{-2}\right), which is the exact value of the integral (approximately 3.6273.627).

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