Question : The area bounded by the curve $y = (x+1)^{2}$, its tangent at point $(1,4)$ and $x = -1$.
I began by finding the tangent to curve at points $(1,4)$
Differentiating the curve $w.r.t.x$ I got : $$y' = 2(x+1)$$
and substituting for $x = 1$, I got the slope of the tangent as : $$m = 4$$
Therefore the equation of the tangent of the curve is: $$y=4x$$
The tangent intersects the $x-axis$ at the origin. And the curve $y = (x+1)^{2}$ intercepts the $x-axis$ at point $-1$.
So to find the area between these 2 curves I integrated as follows:
$$\int_{-1}^{0} (x+1)^{2} – 4x \ =\frac{1}{3}x^{3} + x – x^{2} \ ; from \ x = -1 \ to \ x = 0 $$
$$ = \frac{7}{3} $$
Is this the correct method and solution?
EDIT As mentioned in the comments, integrating between $(-1,1)$ we get:
$$= \frac{8}{3}$$.
Best Answer
Yes your approach is correct but the upper bound of the integral is not correct. It should be $1$ and not $0$. The tangent to the curve is drawn at point $(1,4)$ and note the x-coordinate of the point is $1$. So the correct integral should be,
$ \displaystyle \int_{-1}^{\color {blue} 1} \left[(x+1)^{2} - 4x\right] ~ dx = \frac{8}{3}$