Find the derivative of StartFraction d Over dx EndFraction Integral from 0 to x cubed e Superscript negative t Baseline font siz
e decreased by 3 dt
a. by evaluating the integral and differentiating the result.
b. by differentiating the integral directly.
2 answers:
Answer: (a) e ^ -3x (b)e^-3x
Step-by-step explanation:
I suggest the equation is:
d/dx[integral (e^-3t) dt
First we integrate e^-3tdt
Integral(e ^ -3t dt) as shown in attachment and then we differentiate the result as shown in the attachment.
(b) to differentiate the integral let x = t, and substitute into the expression.
Therefore dx = dt
Hence, d/dx[integral (e ^-3x dx)] = e^-3x
Answer:
(a) 3x²e^(-x³)
(b) -3x²e^(-x³)
Step-by-step explanation:
(a) The integral was evaluated at t = 0 to t = x³
Then the result is differentiated to obtain the result.
(b) The integral is differentiates directly, using the properties of differentiation, the chain rule precisely.
The result obtained in (a) turned out to be the negative of the result obtained in (b)
CHECK ATTACHMENT FOR THE WORKINGS.
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62.3 / 3.5 = 17.8
3,500 m so 3 km and 500 m
L=16
W=21
Set up a systems of equations:
x=length
y=width
xy=336
x+5=y
Use substitution to solve:
x(x+5)=336
x^2+5x=336
Solve using factoring:
x^2+5x-336=0
(x-16)(x+21)=0
x=16 and x= -21
Since length can't be negative, l=16
To find width, plug length into the first equation:
(16)y=336
y=21
So...
L=16
<span>W=21</span>
Answer:
A
Step-by-step explanation: