Question

Difficulty: MediumDefinite Integrals and Area Under Curves

Determine the area under the curve y=3x2−3y = 3x^2 - 3 above the xx-axis between x=1x = 1 and x=3x = 3.

Answer: 20 square units

Answer

The area under the curve between x=1x = 1 and x=3x = 3 is 20 square units.
The area under the curve y=3x2−3y = 3x^2 - 3 from x=1x = 1 to x=3x = 3 is obtained by calculating the definite integral ∫13(3x2−3) dx=[x3−3x]13=(33−3(3))−(13−3(1))=18−(−2)=20\int_{1}^{3} (3x^2 - 3) \, dx = [x^3 - 3x]_{1}^{3} = (3^3 - 3(3)) - (1^3 - 3(1)) = 18 - (-2) = 20.

Step-by-Step Solution

1
Set up the definite integral representing the area under the curve.
A=∫13(3x2−3) dxA = \int_{1}^{3} (3x^2 - 3) \, dx
The area bounded by a non-negative curve y=f(x)y = f(x), the xx-axis, and vertical lines x=ax = a and x=bx = b is given by ∫abf(x) dx\int_{a}^{b} f(x) \, dx.
2
Integrate the polynomial function term by term.
∫(3x2−3) dx=x3−3x+C\int (3x^2 - 3) \, dx = x^3 - 3x + C
Applying the power rule of integration: ∫3x2dx=x3\int 3x^2 dx = x^3 and ∫3dx=3x\int 3 dx = 3x.
3
Apply the Fundamental Theorem of Calculus by substituting the limits of integration.
[x3−3x]13=(33−3(3))−(13−3(1))=18−(−2)=20[x^3 - 3x]_{1}^{3} = (3^3 - 3(3)) - (1^3 - 3(1)) = 18 - (-2) = 20
Evaluating F(b)−F(a)F(b) - F(a) gives (27−9)−(1−3)=18−(−2)=20(27 - 9) - (1 - 3) = 18 - (-2) = 20.

Key Concept

Definite Integrals and Area Under Curves
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