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Evaluate the definite integral int−22(3x3−2x+1),dxint_{-2}^2 (3x^3 - 2x + 1) , dx.
Aug 14, 2024
Evaluate the definite integral int−22(3x3−2x+1),dxint_{-2}^2 (3x^3 - 2x + 1) , dx.
Generated Graph
Solution by Steps
step 1
We need to evaluate the definite integral ∫−22(3x3−2x+1) dx \int_{-2}^{2} (3x^3 - 2x + 1) \, dx
step 2
The integral can be computed as follows: ∫(3x3−2x+1) dx=3x44−x2+x+C \int (3x^3 - 2x + 1) \, dx = \frac{3x^4}{4} - x^2 + x + C
step 3
Now, we evaluate the definite integral from -2 to 2: [3(2)44−(2)2+(2)]−[3(−2)44−(−2)2+(−2)] \left[ \frac{3(2)^4}{4} - (2)^2 + (2) \right] - \left[ \frac{3(-2)^4}{4} - (-2)^2 + (-2) \right]
step 4
Calculating the values: [3(16)4−4+2]−[3(16)4−4−2]=(12−4+2)−(12−4−2)=10−6=4 \left[ \frac{3(16)}{4} - 4 + 2 \right] - \left[ \frac{3(16)}{4} - 4 - 2 \right] = (12 - 4 + 2) - (12 - 4 - 2) = 10 - 6 = 4
Answer
4
Key Concept
Definite integrals represent the area under a curve between two points.
Explanation
The result of the definite integral gives the net area between the curve 3x3−2x+13x^3 - 2x + 1 and the x-axis from x=−2x = -2 to x=2x = 2, which is calculated to be 4.
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