Question 1

Q1: Find the area of the region bounded by the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$.

Accepted Answer

Given: The equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$. To Find: The area of the region bounded by the ellipse. Solution: The given equation of the ellipse is $\frac{x^{2}}{4^2} + \frac{y^{2}}{3^2} = 1$. This is a standard ellipse with semi-major axis $a=4$ and semi-minor axis $b...

Question 2

Q2: Find the area of the region bounded by the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$.

Accepted Answer

Given: The equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$. To Find: The area of the region bounded by the ellipse. Solution: The given equation of the ellipse is $\frac{x^{2}}{2^2} + \frac{y^{2}}{3^2} = 1$. This is a standard ellipse with semi-major axis $b=3$ along the y-axis and se...

Question 3

Q3: Choose the correct answer in the following Exercises 3 and 4. 3. Area lying in the first quadrant and bounded by the circle $x^{2}+y^{2}=4$ and the li...

Accepted Answer

Given: The equation of the circle is $x^2 + y^2 = 4$, and the lines are $x=0$ and $x=2$. To Find: The area of the region in the first quadrant bounded by the circle and the given lines. Solution: The equation of the circle can be written as $x^2 + y^2 = 2^2$. This is a circle with its center at the...

Question 4

Q4: Area of the region bounded by the curve $y^{2}=4 x, y$-axis and the line $y=3$ is (A) 2 (B) $\frac{9}{4}$ (C) $\frac{9}{3}$ (D) $\frac{9}{2}$

Accepted Answer

Given: The curve is $y^2 = 4x$, bounded by the y-axis and the line $y=3$. To Find: The area of the specified region. Solution: The curve is a parabola with its vertex at the origin, opening to the right. The region is bounded by the y-axis ($x=0$), the line $y=3$, and the parabola. Since the boundar...

Question 5

Q1: Find the area under the given curves and given lines: (i) $y=x^{2}, x=1, x=2$ and $x$-axis (ii) $y=x^{4}, x=1, x=5$ and $x$-axis

Accepted Answer

Part (i): Given: The curve is $y=x^2$, bounded by the lines $x=1$, $x=2$ and the x-axis. To Find: The area of the specified region. Solution: The curve $y=x^2$ is a parabola opening upwards. In the interval $[1, 2]$, the curve is above the x-axis. The area is given by the definite integral of $y$ wi...

Question 6

Q2: Sketch the graph of $y=|x+3|$ and evaluate $\int_{-6}^{0}|x+3| d x$.

Accepted Answer

Given: The function $y=|x+3|$ and the integral $\int_{-6}^{0}|x+3| d x$. To Find: Sketch the graph and evaluate the integral. 1. Sketching the Graph: The function is $y = |x+3|$. The graph of $y=|x|$ is a V-shape with its vertex at the origin. The graph of $y=|x+3|$ is a translation of $y=|x|$ by 3...

Question 7

Q3: Find the area bounded by the curve $y=\sin x$ between $x=0$ and $x=2 \pi$.

Accepted Answer

Given: The curve is $y=\sin x$, bounded between $x=0$ and $x=2\pi$. To Find: The area of the specified region. Solution: The graph of $y=\sin x$ is above the x-axis in the interval $[0, \pi]$ and below the x-axis in the interval $[\pi, 2\pi]$. To find the total area, we must calculate the area of ea...

Question 8

Q4: Choose the correct answer in the following Exercises from 4 to 5. 4. Area bounded by the curve $y=x^{3}$, the $x$-axis and the ordinates $x=-2$ and $x...

Accepted Answer

Given: The curve is $y=x^3$, bounded by the x-axis and the lines $x=-2$ and $x=1$. To Find: The area of the specified region. Solution: The graph of $y=x^3$ is below the x-axis for $x  0$. The interval of integration is $[-2, 1]$. We must split the integral at $x=0$ to calculate the area correctly,...

Question 9

Q5: The area bounded by the curve $y=x|x|, x$-axis and the ordinates $x=-1$ and $x=1$ is given by (A) 0 (B) $\frac{1}{3}$ (C) $\frac{2}{3}$ (D) $\frac{4}{...

Accepted Answer

Given: The curve is $y=x|x|$, bounded by the x-axis and the lines $x=-1$ and $x=1$. To Find: The area of the specified region. Solution: First, we define the function $y=x|x|$ piecewise: $y = \begin{cases} x(x) = x^2 & 	ext{if } x \geq 0 \ x(-x) = -x^2 & 	ext{if } x < 0 \end{cases}$ The interval o...

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