What is the area bounded by the curves y=4x^2-25, y=0, x=-1, x=3.5? PS the area is not 54 or 60 I already tried.

Answer 1

#76.67# ( 2 .d.p.)

#f(x)=4x^2-25#

Find roots:

#4x^2-25=0#

#x=sqrt(25/4)=>x=5/2 and -5/2#

So our given area lies below and above the x axis.

It is below in the interval #[-1,5/2]#

and above in the interval #[5/2, 7/2]#

So we require the integrals:

#int_(-1)^(5/2)(4x^2-25)dxcolor(white)(88)# and #color(white)(88)int_(5/2)^(7/2)(4x^2-25)dx#

1st:

#"Area" =int_(-1)^(5/2)(4x^2-25)dx=4/3x^3-25x=[4/3x^3-25x]_(-1)^(5/2)#

#=[4/3x^3-25x]^(5/2)-[4/3x^3-25x]_(-1)#

Plugging in upper and lower bounds:

#=[4/3(5/2)^3-25(5/2)]^(5/2)-[4/3(-1)^3-25(-1)]_(-1)#

#=[125/6-125/2]^(5/2)-[-4/3+25]_(-1)#

#=[-125/3]^(5/2)-[71/3]_(-1)#

#=-125/3-71/3=-196/3#

#"Area" = 196/3#

2nd:

#"Area" =int_(5/2)^(7/2)(4x^2-25)dx=4/3x^3-25x#

#=[4/3x^3-25x]_(5/2)^(7/2)#

#=[4/3x^3-25x]^(7/2)-[4/3x^3-25x]_(5/2)#

Plugging in upper and lower bounds:

#=[4/3(7/2)^3-25(7/2)]^(7/2)-[4/3(5/2)^3-25(5/2)]_(5/2)#

#=[343/6-175/2]^(7/2)-[125/6-125/2]_(5/2)#

#=-91/3-(-125/3)=34/3#

#"Area = 34/3#

Total area #= 196/3+34/3=230/3=76.67# ( 2 .d.p.)

GRAPH:

You probably forgot to remove the negation when adding the two areas.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To find the area bounded by the curves y = 4x^2 - 25, y = 0, x = -1, and x = 3.5, you need to integrate the function (4x^2 - 25) between the given limits of integration, -1 and 3.5, and then take the absolute value of the result. This is because the curve may dip below the x-axis.

So, the integral you need to solve is:

[ A = \left| \int_{-1}^{3.5} (4x^2 - 25) , dx \right| ]

[ A = \left| \left[ \frac{4}{3}x^3 - 25x \right]_{-1}^{3.5} \right| ]

[ A = \left| \left( \frac{4}{3}(3.5)^3 - 25(3.5) \right) - \left( \frac{4}{3}(-1)^3 - 25(-1) \right) \right| ]

[ A = \left| \left( \frac{4}{3}(42.875) - 87.5 \right) - \left( \frac{4}{3}(-1) + 25 \right) \right| ]

[ A = \left| \left( 57.166 - 87.5 \right) - \left( -\frac{4}{3} + 25 \right) \right| ]

[ A = \left| \left( -30.334 \right) - \left( 23.666 \right) \right| ]

[ A = \left| -54 \right| ]

So, the area bounded by the curves (y = 4x^2 - 25), (y = 0), (x = -1), and (x = 3.5) is (54).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
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.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7