How do you integrate $\int 2^{x} d x$ $int 2^xdx$ from $[- 1, 2]$ $[-1,2]$ ?

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Answer 1 · 2016-12-15T14:11:55

The answer is $= \frac{7}{2 ln 2} = 5.05$ $=7/(2ln2)=5.05$

Let $u = 2^{x}$ $u=2^x$

$ln u = x ln 2$ $lnu=xln2$

$u = e^{x ln 2}$ $u=e^(xln2)$

$\int 2^{x} d x = \int e^{x ln 2} d x = \frac{e^{x ln 2}}{ln 2} = \frac{2^{x}}{ln 2}$ $int2^xdx=inte^(xln2)dx=e^(xln2)/ln2=2^x/ln2$

$\int_{- 1}^{2} 2^{x} d x = {[\frac{2^{x}}{ln 2}]}_{- 1}^{2}$ $int_-1^2 2^xdx= [2^x/ln2] _-1^2$

$= \frac{1}{ln 2} (2^{2} - \frac{1}{2})$ $=1/ln2(2^2-1/2)$

$= \frac{1}{ln 2} \cdot \frac{7}{2} = 5.05$ $=1/ln2*7/2=5.05$

How do you integrate ∫2xdxint 2^xdx from [−1,2][-1,2]?