Sketch the graph of y=|x+3| and evaluate int_ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given equation is $y=|x+3|$.The corresponding values of $x$ and $y$ are given in the following table:$X$$-6$$-5$$-4$$-3$$-2$$-1$$0$$Y$$3$$2$$1

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) The given equation is $y=|x+3|$.The corresponding values of $x$ and $y$ are given in the following table:$X$$-6$$-5$$-4$$-3$$-2$$-1$$0$$Y$$3$$2$$1$$0$$1$$2$$3$On plotting these points,we obtain the graph of $y=|x+3|$.It is known that $(x+3) \leq 0$ for $-6 \leq x \leq -3$ and $(x+3) \geq 0$ for $-3 \leq x \leq 0$.Therefore,$\int_{-6}^{0}|x+3| d x = \int_{-6}^{-3}-(x+3) d x + \int_{-3}^{0}(x+3) d x$.$= -\left[\frac{x^2}{2} + 3x\right]_{-6}^{-3} + \left[\frac{x^2}{2} + 3x\right]_{-3}^{0}$.$= -\left[\left(\frac{9}{2} - 9\right) - \left(\frac{36}{2} - 18\right)\right] + \left[(0) - \left(\frac{9}{2} - 9\right)\right]$.$= -\left[-\frac{9}{2} - 0\right] + \left[0 - (-\frac{9}{2})\right]$.$= \frac{9}{2} + \frac{9}{2} = 9$.

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

Similar Questions

The area (in sq. units) of the smaller part of the circle $x^2+y^2=a^2$ cut off by the line $x=\frac{a}{\sqrt{2}}$ is

The area bounded by the curve $x=4-y^{2}$ and the $Y$-axis is

The area of the region bounded by the curve $y^2 = x^3$, the $y$-axis and the lines $y = 1$ and $y = 8$ is

The area of the region (in sq. units) bounded by the curve $y = 2\sqrt{1 - x^2}$ and the $X$-axis is . . . . . . .

Find the area of the circle $4x^{2}+4y^{2}=9$ which is interior to the parabola $x^{2}=4y$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module