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attashe74 [19]
1 year ago
11

5.3 Q4

Mathematics
1 answer:
barxatty [35]1 year ago
8 0

The derivative of the function g(x) as given in the task content by virtue of the Fundamental theorem of calculus is; g'(x) = √2 ln(t) dt = 1.

<h3>What is the derivative of the function g(x) by virtue of the Fundamental theorem of calculus as given in the task content?</h3>

g(x) = Integral; √2 ln(t) dt (with the upper and lower limits e^x and 1 respectively).

Since, it follows from the Fundamental theorem of calculus that given an integral where;

Now, g(x) = Integral f(t) dt with limits a and x, it follows that the differential of g(x);

g'(x) = f(x).

Consequently, the function g'(x) which is to be evaluated in this scenario can be determined as:

g'(x) = \int\limits^{e^x}_1 2 ln(t) dt = 1

The derivative of the function g(x) as given in the task content by virtue of the Fundamental theorem of calculus is; g'(x) = √2 ln(t) dt = 1.

Read more on fundamental theorem of calculus;

brainly.com/question/14350153

#SPJ1

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b_1=\frac{2A}{h}-b_2

Step-by-step explanation:

Given expression is;

A=\frac{h(b_1+b_2)}{2}

Solving for b_1

Multiplying both sides by 2

2*A=\frac{h(b_1+b_2)}{2}*2\\2A=h(b_1+b_2)

Dividing both sides by h

\frac{2A}{h}=\frac{h(b_1+b_2)}{h}\\\frac{2A}{h}=b_1+b_2

Subtracting b_2 from both sides

\frac{2A}{h}-b_2=b_1+b_2-b_2\\\frac{2A}{h}-b_2=b_1\\b_1=\frac{2A}{h}-b_2

b_1=\frac{2A}{h}-b_2

Keywords: division, subtraction

Learn more about subtraction at:

  • brainly.com/question/11145277
  • brainly.com/question/11150876

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The answer is 6, 9, 13.5
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I really need this answered fast.....The coordinates of the vertices of △JKL are J(−5, −1) , K(0, 1) , and L(2, −5) . Which stat
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For the function f(x)=(x+6)^3, rind f^-1(x)
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Answer:

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Step-by-step explanation:

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y=(x+6)^3

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