2017年2月17日 · Note that #16=2^4#. #rArr2^(x+2)=(2^4)^x# #rArrcolor(red)(2)^(x+2)=color(red)(2)^(4x)# Since the bases are equal, that is #color(red)(2)# then the exponents are equal.

2018年3月23日 · Perform the substitution. #x=4sintheta#, #=>#, #dx=4costhetad theta# #sqrt(16-x^2)=sqrt(16-16sin^2theta)=4costheta#

2018年6月1日 · Substitute #x=4sin theta# to turn the denominator into a simple trig function via the identity #sin^2 theta +cos^2 theta=1#. Note that #(dx)/(d theta)=4 cos theta# and that the limits of integration #x=0,2sqrt(3)# are equivalent to #theta=0,pi/3# .

c = 16 Substitute these values into the formula above to find that (0, 16) is the vertex point. The expression 16 - x^2 can be factored as a difference of squares. (4 + x)*(4 - x)=0 So, the x-intercepts of the graph are (-4, 0) and (4, 0). Pick some other values of x, and find the corresponding values of y to get a few more points to plot.

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2016年9月10日 · -sqrt(16-x^2)+C Although this is well set up for a shorter, non-trigonometric substitution (see: u=16-x^2), we can make the trigonometric substitution x=4sintheta. Note that this means that dx=4costhetad theta.

2018年3月11日 · I find that problems like this are solved easier by applying our rules of logarithmic functions. #y = x*sqrt(16-x^2)# We know from the rules of algebra that manipulated one side of an equation by applying an operator to both sides is still an equivalent function, but in an alternative form.

The critical numbers are +-2sqrt2, and +-4. Critical numbers for a function, f, are numbers in the domain f at which f'(x) = 0 or f'(x) does not exist. For, f(x) = xsqrt(16 - x^2), the derivative is: f'(x) = sqrt(16-x^2) + x 1/(2sqrt(16-x^2)) *(-2x) = sqrt(16-x^2)/1 - x^2/sqrt(16-x^2) = (16-2x^2)/sqrt(16-x^2) Derivative is 0 f'(x) = 0 when 16-2x^2 = 0. Which happens at x=+-sqrt8 = +-2sqrt2 ...

2018年5月5日 · Here, #I=intx^3sqrt(16-x^2)dx# Let, #x=4sinu=>dx=4cosudu# #and sinu=x/4=>cosu=sqrt(1-sin^2u)=sqrt(1-x^2/16# ...

2018年1月28日 · Try #x=4 tan theta#.Then #x = 4 sec^2theta d theta# So #int x^3/sqrt{x^2+16} dx = int {64 tan^3 theta times 4 sec^2theta}/{4sectheta}d theta #