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

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(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

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

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(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

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