How do you evaluate the definite integral #int -8/3*2^xdx# from -2 to 9?

Question

Emilia Aguilar · Accepted Answer

#-4094/(3ln2)#

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If you already know the formula for integrals involving #bb(a^x)#:

We can use the formula: #inta^xdx=1/lnaa^x+C#

So, we see that:

#int_(-2)^9-8/3*2^xdx=-8/3int_(-2)^9 2^xdx=-8/(3ln2)[2^x]_(-2)^9#

Evaluating, this becomes:

#-8/(3ln2)[2^x]_(-2)^9=-8/(3ln2)(2^9-2^(-2))=-8/(3ln2)(512-1/4)#

#=-8/(3ln2)(2047/4)=-4094/(3ln2)#

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Without knowing the formula:

You should know that #inte^udu=e^u+C#. We can use this to our advantage here, since #2^x=e^(ln(2^x))=e^(xln2)#. Thus:

#-8/3int_(-2)^9 2^xdx=-8/3int_(-2)^9e^(xln2)dx#

Here, let #u=xln2#. This implies that #du=ln2*dx#. So, multiply the integrand by #ln2# and the exterior by #1/ln2#.

#=-8/(3ln2)int_(-2)^9 ln2*e^(xln2)dx#

Before we make the #u# and #du# substitutions, recall that we will have to switch our bounds by plugging them in for #x# in #u=xln2#. Thus #-2->-2ln2# and #9rarr9ln2#.

#=-8/(3ln2)int_(-2ln2)^(9ln2)e^udu=-8/(3ln2)[e^u]_(-2ln2)^(9ln2)#

From here, the #e# and #ln# will cancel out (remember to write #-2ln2# as #ln(2^(-2))=ln(1/4)# and #9ln2# as #ln(2^9)=ln512#), and then the work becomes the same as the method outlined above.

Paisley Johnson · Answer

undefined To evaluate the definite integral $ \int_{-2}^{9} -\frac{8}{3} \cdot 2^x \, dx $, we can use the properties of exponential functions and the definite integral.

First, we'll find the antiderivative of $ -\frac{8}{3} \cdot 2^x $. The antiderivative of $ 2^x $ is $ \frac{1}{\ln(2)} \cdot 2^x $. So, the antiderivative of $ -\frac{8}{3} \cdot 2^x $ is $ -\frac{8}{3\ln(2)} \cdot 2^x $.

Now, we'll evaluate the definite integral using the antiderivative:

$$ \int_{-2}^{9} -\frac{8}{3} \cdot 2^x \, dx = \left[ -\frac{8}{3\ln(2)} \cdot 2^x \right]_{-2}^{9} $$

Plugging in the upper and lower limits:

$$ = -\frac{8}{3\ln(2)} \cdot 2^9 - \left( -\frac{8}{3\ln(2)} \cdot 2^{-2} \right) $$

$$ = -\frac{8}{3\ln(2)} \cdot 512 + \frac{8}{3\ln(2)} \cdot \frac{1}{4} $$

$$ = -\frac{8}{3\ln(2)} \cdot 512 + \frac{2}{3\ln(2)} $$

$$ = -\frac{8 \cdot 512}{3\ln(2)} + \frac{2}{3\ln(2)} $$

$$ = -\frac{4096}{\ln(2)} + \frac{2}{3\ln(2)} $$

$$ = -\frac{4096 - 2}{\ln(2)} $$

$$ = -\frac{4094}{\ln(2)} $$

So, $ \int_{-2}^{9} -\frac{8}{3} \cdot 2^x \, dx = -\frac{4094}{\ln(2)} $.