How to answer these using intergration ?

Answer 1

The area is #=(32/3)u^2# and the volume is #=(512/15pi)u^3#

Start by finding the intercept with the x-axis

#y=4x-x^2=x(4-x)=0#

Therefore,

#x=0# and #x=4#

The area is

#dA=ydx#
#A=int_0^4(4x-x^2)dx#
#=[2x^2-1/3x^3]_0^4#
#=32-64/3-0#
#=32/3u^2#

The volume is

#dV=piy^2dx#
#V=piint_0^4(4x-x^2)^2dx#
#=piint_0^4(16x^2-8x^3+x^4)dx#
#=pi[16/3x^3-2x^4+1/5x^5]_0^4#
#=pi(1024/3-512+1024/5-0)#
#=pi(5120/15-7680/15+3072/15)#
#=pi(512/15)#
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Answer 2

a. #32/3#

b. #(512pi)/15#

First, we need to find the points at which the graph crosses the #x#-axis.
#4x-x^2=x(4-x)=0#
Either #x=0# or #4-x=0#
#x=0 or 4#

Now we know our upper and lower bounds.

a. #"Area under a graph"=int_b^af(x)dx#
#int_0^4 4x-x^2dx=[2x^2-x^3/3]_0^4=(2(4)^2-4^3/3)-(2(0)^2-0^3/3)=32/3#
b. #"Volume of rotation"=piint_b^a(f(x))^2dx#
#f(x)^2=(4x-x^2)^2=16x^2-8x^3+x^4#
#piint_0^4 16x^2-8x^3+x^4dx=pi[(16x^3)/3-2x^4+x^5/5]_0^4=pi[((16(4)^3)/3-2(4)^4+4^5/5)-((16(0)^3)/3-2(0)^4+0^5/5)]=pi[512/15]=(512pi)/15#
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Answer 3

To answer questions using integration, you need to follow these steps:

  1. Understand the problem and identify what needs to be integrated.
  2. Set up the integral using appropriate notation.
  3. Evaluate the integral using integration techniques such as substitution, integration by parts, trigonometric substitutions, or partial fractions.
  4. Check your solution by differentiating the result and ensuring it matches the original function, if applicable.
  5. Simplify the answer as much as possible, and include any necessary constants of integration.

If you have specific questions or examples, feel free to ask for further clarification or guidance.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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