The area (in sq. units) bounded by the curve | vedclass

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

Question

(A) The equation of the curve is $y = x^2 + 2x + 1 = (x + 1)^2$.First,we find the equation of the tangent at $(1, 4)$.Differentiating $y$ with respect

Vedclass · Accepted Answer

$\frac{1}{3} \ sq. \ units$

Vedclass · Answer

(A) The equation of the curve is $y = x^2 + 2x + 1 = (x + 1)^2$.First,we find the equation of the tangent at $(1, 4)$.Differentiating $y$ with respect to $x$,we get $\frac{dy}{dx} = 2x + 2$.At the point $(1, 4)$,the slope of the tangent is $m = 2(1) + 2 = 4$.The equation of the tangent line is $y - 4 = 4(x - 1)$,which simplifies to $y = 4x$.The area bounded by the curve,the tangent,and the $Y$-axis is the area under the curve from $x = 0$ to $x = 1$ minus the area of the triangle formed by the tangent line,the $X$-axis,and the vertical line $x = 1$.Area $A = \int_0^1 (x^2 + 2x + 1) dx - \int_0^1 (4x) dx$.Calculating the first integral: $\int_0^1 (x^2 + 2x + 1) dx = \left[ \frac{x^3}{3} + x^2 + x \right]_0^1 = \frac{1}{3} + 1 + 1 = \frac{7}{3}$.Calculating the second integral (area of the triangle): $\int_0^1 4x dx = \left[ 2x^2 \right]_0^1 = 2(1)^2 - 0 = 2$.Therefore,the required area $A = \frac{7}{3} - 2 = \frac{7 - 6}{3} = \frac{1}{3} \ sq. \ units$.

The area (in $sq. \ units$) bounded by the curve $y = x^2 + 2x + 1$,the tangent to it at $(1, 4)$,and the $Y$-axis is

Similar Questions

Using integration,find the area of the triangular region whose sides have the equations $y=2x+1$,$y=3x+1$,and $x=4$.

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

Find the area under the given curve $y=x^{2}$ bounded by the lines $x=1$,$x=2$ and the $x$-axis.

The area bounded by the curve $y = x(1 - \ln x)$,the line $x = e^{-1}$,and the positive $X$-axis between $x = e^{-1}$ and $x = e$ is:

$AOB$ is the positive quadrant of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ in which $OA=5, OB=3$. The area between the arc $AB$ and the chord $AB$ of the ellipse in sq. units is

Vedclass Test Series

Exam Paper Generator

Online Exam Module