The area bounded by the two graphs above is equal to:

Question

BlackTom AI · Accepted Answer

Solve the intersection first: set x^2 = 2x + 3 to find the interval of integration. This gives
x^2 - 2x - 3 = 0 => (x - 3)(x + 1) = 0, so x = -1 and x = 3.

1) Relevant knowledge points and concepts
- Area between two curves y = f(x) and y = g(x) on [a,b] is ∫_a^b (top function − bottom function) dx.
- Determine which function is on top by testing a value between the intersection points.

2) Step-by-step reasoning
- Test x = 0 (between -1 and 3): y_parabola = 0^2 = 0, y_line = 2(0)+3 = 3. The line y = 2x + 3 lies above the parabola y = x^2 on the entire interval [-1,3].
- So area = ∫_{-1}^{3} [(2x + 3) − x^2] dx.

3) Calculations with intermediate steps
Compute the antiderivative:
∫(2x + 3 − x^2) dx = ∫2x dx + ∫3 dx − ∫x^2 dx
= x^2 + 3x − (x^3/3) + C.

Evaluate from -1 to 3:
F(3) = 3^2 + 3(3) − (3^3/3) = 9 + 9 − 27/3 = 9 + 9 − 9 = 9.
F(−1) = (−1)^2 + 3(−1) − ((−1)^3/3) = 1 − 3 − (−1/3) = 1 − 3 + 1/3 = −2 + 1/3 = −5/3.

Definite integral = F(3) − F(−1) = 9 − (−5/3) = 9 + 5/3 = (27/3 + 5/3) = 32/3 = 10 2/3.

4) Why the correct answer is correct
The calculation follows directly from the intersection points and the definite integral of the vertical distance (line minus parabola). All algebra and arithmetic reduce to 32/3 square units, which as a mixed number is 10 2/3.

5) Why the other options are incorrect
- a. 7 1/3 = 22/3, not equal to 32/3.
- b. 6 2/3 = 20/3, not equal to 32/3.
- d. 11 1/3 = 34/3, not equal to 32/3.
None of these match the evaluated integral.

Conclusion: The area bounded by y = x^2 and y = 2x + 3 is 32/3 = 10 2/3 square units, which matches option c.

The area bounded by the two graphs above is equal to:Single choice

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