How to Calculate Area in a Graph Using Integrals?

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SUMMARY

The discussion focuses on calculating the area under the curve defined by the function Y = X² + 4X + 5 over the interval -1 ≤ X ≤ 0. The correct integral setup involves using the antiderivative F(x) = (1/3)x³ + 2x² + 5x, with the area calculated as Area = F(0) - F(-1). The user incorrectly computed F(-1) as -3.5 and misunderstood the limits of integration, leading to confusion in the area calculation.

PREREQUISITES
  • Understanding of definite integrals
  • Familiarity with antiderivatives and the Fundamental Theorem of Calculus
  • Knowledge of polynomial functions and their properties
  • Basic skills in calculus, specifically integration techniques
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  • Review the Fundamental Theorem of Calculus
  • Practice calculating definite integrals with polynomial functions
  • Learn how to correctly apply limits of integration in area calculations
  • Explore common mistakes in integral calculus to avoid similar errors
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Students and educators in calculus, particularly those focusing on integral calculus and area calculations under curves. This discussion is beneficial for anyone seeking to clarify misconceptions about integration limits and antiderivatives.

SwedishFred
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Calculate area D=(x,y): -1≤X≤0 0≤Y≤ X²+4x+5

I started with dA=f(x) dx
∫f(Y=x²+4x+5) [F(x) x^3/3 + 2X²+5X] higer limit 0 lower limit -1

F(0)=0
F(-1)=-3.5
F(a)-F(b) = -3,5

I don't get this ... ??
What am i missing?

Regards!
 
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SwedishFred said:
Calculate area D=(x,y): -1≤X≤0 0≤Y≤ X²+4x+5

I started with dA=f(x) dx
∫f(Y=x²+4x+5) [F(x) x^3/3 + 2X²+5X] higer limit 0 lower limit -1

F(0)=0
F(-1)=-3.5
F(a)-F(b) = -3,5

I don't get this ... ??
What am i missing?

Regards!

1) For ##F(x) = \frac{1}{3}x^3 + 2 x^2 + 5x## we have ##F(-1) \neq -3.5##.
2) ##\text{Area} = F(0) - F(-1)##, not the other way round.
 
Also, even if F(-1) happened to be equal to -3.5 (it isn't), 0 - (-3.5) ≠ -3.5.

One more thing - questions about derivatives or integrals should be posted in the Calculus & Beyond section, not in the Precalculus section.
 

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