The region is bounded by the parabola $y^2=4x$, the X-axis ($y=0$), $x=0$, and $x=1$.
Since the parabola is symmetric about the X-axis, the total area can be found by calculating the area of the upper half (where $y = 2\sqrt{x}$) and multiplying by 2.
1. From $y^2=4x$, we get $y = \pm \sqrt{4x} = \pm 2\sqrt{x}$.
2. The area $A$ is given by the integral of $y$ with respect to $x$ from $x=0$ to $x=1$. Due to symmetry, we calculate the area for $y=2\sqrt{x}$ and multiply by 2.
$$A = 2 \int_{0}^{1} 2\sqrt{x} \, dx$$
3. Evaluate the integral:
$$A = 4 \int_{0}^{1} x^{1/2} \, dx$$
$$A = 4 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1}$$
$$A = 4 \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1}$$
4. Apply the limits of integration:
$$A = \frac{8}{3} \left( (1)^{3/2} - (0)^{3/2} \right)$$
$$A = \frac{8}{3} (1 - 0)$$
$$A = \frac{8}{3}$$
The final answer is $\boxed{\frac{8}{3}}$.
AI Tutor Explanation
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Step-by-Step Solution
The equation of the curve is \(y^2 = 4x\). We need to find the area bounded by this curve, the X-axis, and the lines \(x=0\) and \(x=1\).
First, express \(y\) in terms of \(x\):
\(y = \pm \sqrt{4x} = \pm 2\sqrt{x}\)
Since we are considering the area bounded by the X-axis, we will take the positive part of \(y\), i.e., \(y = 2\sqrt{x}\).
The area \(A\) can be found by integrating \(y\) with respect to \(x\) from \(x=0\) to \(x=1\):
\(A = \int_{0}^{1} 2\sqrt{x} \, dx\)
Now, evaluate the integral:
\(A = 2 \int_{0}^{1} x^{\frac{1}{2}} \, dx = 2 \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right]_{0}^{1} = 2 \cdot \frac{2}{3} \left[ x^{\frac{3}{2}} \right]_{0}^{1}\)
\(A = \frac{4}{3} \left[ 1^{\frac{3}{2}} - 0^{\frac{3}{2}} \right] = \frac{4}{3} \left[ 1 - 0 \right] = \frac{4}{3}\)
Correct Answer: \(\frac{4}{3}\)<\/strong>
Pedagogical Audit
Bloom's Analysis:
This is an APPLY question because it requires the student to apply the concept of definite integrals to calculate the area under a curve. The student needs to know the formula for area calculation using integration and apply it to the given function and limits.
Knowledge Dimension:PROCEDURAL
Justification:The question requires the student to follow a specific procedure (integration) to find the area. It involves knowing the steps of integration and applying them correctly.<\/span>
Syllabus Audit:
In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of applying integration to find the area under a curve, a standard topic covered in the textbook.
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