Finding the area between a line and a curve

Question

The two equations are $x+1$ and $4x-x^2-1$.

The answer is $\frac{1}{6}$, but I've done it 4 different times and gotten -$\frac{15}{2}$ each time.

My working:

1. $x+1$ = $4x-x^2-1$
2. $x^2-3x+2 = 0$
3. $(x-1)(x-2)$ means $x=1$ or $x=2$
4. $\int_1^2$ $3x-x^2$
5. $[\frac{3x^2}{2}-\frac{x^3}{3}]_1^2$
6. $\frac{3(1)^2}{2}-\frac{(1)^3}{3}$ = $\frac{3}{2}-\frac{1}{3}$
7. $\frac{3(2)^2}{2}-\frac{(2)^3}{3}$ = $\frac{12}{2}-\frac{8}{3}$
8. ($\frac{3}{2}-\frac{1}{3}$)-($\frac{12}{2}-\frac{8}{3}$) = $-\frac{9}{2}-\frac{9}{3}$
9. -$\frac{15}{2}$

Answer 1

You got the two $x$ points right, but you make a mistake in the next step. The correct version should be:

$$\int_1^2[\text{top function}-\text{bottom function}]\ dx$$

Here, the top function is $4x-x^2-1$ and the bottom function is $x+1$, hence

$$\begin{align}\text{Area}&=\int_1^2[(4x-x^2-1)-(x+1)]\ dx\\&=\int_1^2[4x-x^2-1-x-1]\ dx\\&=\int_1^2[3x-x^2-2]\ dx\end{align}$$

Can you take it from here?

Answer 2

So the area is $$\begin{align}A&=\int_{1}^2\big[(4x-x^2-1)-(x+1)\big]dx\\ &=\int_{1}^2(-x^2+3x-2)dx\\ &=\Big[\frac{-x^3}{3}+\frac{3x^2}{2}-2x\Big]_{1}^{2}\\ &=\left(-\frac{8}{3}+6-4\right)-\left(-\frac{1}{3}+\frac{3}{2}-2\right)\\ &=-\frac{8}{3}+2+\frac{1}{3}-\frac{3}{2}+2\\ &=-\frac{7}{3}+4-\frac{3}{2}\\ &=-\frac{7}{3}+\frac{5}{2}\\ &=\frac{1}{6}\\ \end{align}$$

Edit: I got down voted. Well, there is nothing I can do. My only point is to show to the OP that his integral $$\int_{1}^2(3x-x^2)dx$$ is wrong. Maybe he thought that $1$ cancels out. So, I showed how he should have the correct computations.

Answer 3

Hint: You need to be doing top curve minus bottom curve i.e;

$$(4x-x^2-1) - (x+1) = 3x-x^2-2$$

Answer 4

There is a mistake in the line below:

5. $[\frac{3x^2}{2}-\frac{x^3}{3}]_1^2$

Actually you should be integrating the difference of the $2$ curves within that limit, i.e. $(4x-x^2-1)-(x+1)=3x-x^2-2$, since each of them represents the area under the curve and bounded by the x-axis.

The magnitude of the enclosed area will, accordingly, be $$\begin{align} & \int_1^2 \{(4x-x^2-1)-(x+1)\} \mathrm{dx} \\ & =\int_1^2 (3x-x^2-2) \mathrm{dx} \\ & =\left[\frac{3x^2}{2}-\frac{x^3}{3}-2x\right]_1^2 \\ & =\frac{3(2^2-1)}{2}-\frac{2^3-1}{3}-2(2-1) \\ & =\frac{9}{2}-\frac{7}{3}-2 \\ & =\frac{27-14-12}{6} \\ & =\color{blue}{\frac{1}{6}}\end{align}$$

Answer 5

Step 5. onwards is not ok

$$\int_a^bf(x) dx = F(b)-F(a) $$

$$\int_a^bf(x) dx = F(b-a) $$

The first result is correct but second one is in general wrong.

EDIT1:

The constant term in $ (x^2-3 x +2) $ is missing for integration. Factorization is not beneficial,