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When you evaluate a definite integral, we can think about using the "+C" and see what happens. Let's take (INT)2x dx between 2 and 3. We then have [x2+C] between 2 and 3. For x=3 we h...
Differentiate x = u2 to get dx = 2u du. We need to change the limits, too:
1 <= x <= 9 <==> 1 <= u2 <= 9 <==> 1 <= u <= 3 (since we are give...
We want to prove:
alogb(c) = clogb(a).
Recall that we can always write x = eln(x), so xy = (eln(x))yAnswered by Tutor69809 D. • Maths tutor5135 Views
We will call the integral I, so I = integral of 1/(3+(x+4)1/2) dx between 0 and 5. First substitute u=3+(x+4)1/2 into the equation to get I = integral of 1/u dx between 0 and 5 Next ...
a) 2ln(2x + 1)-4=0
-> ln(2x + 1)-2=0
-> ln(2x + 1)=2
-> (2x + 1)=e2
-> 2x = e2 -1
-> x = (e2 -1)/2
b) 7xAnswered by Elizabeth L. • Maths tutor4264 Views
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