Work — Question 1

Question 1 – Solution

1. Identify the force and interval.

The force acts along the displacement and is $F(x)=3x^2+2x$ newtons. The object moves from $x=0$ to $x=2$ meters.

2. Write the work integral.

Over a small displacement $dx$, $dW=F(x)\,dx$. Therefore,$$W=\int_0^2F(x)\,dx=\int_0^2(3x^2+2x)\,dx.$$

3. Integrate term by term.

By the power rule,$$\int3x^2\,dx=3\frac{x^3}{3}=x^3,\qquad\int2x\,dx=2\frac{x^2}{2}=x^2.$$

4. Apply the bounds.

$$W=\left[x^3+x^2\right]_0^2=(2^3+2^2)-(0^3+0^2)=8+4=12.$$

5. State the result with units.

$$\boxed{W=12\ \mathrm{J}}.$$A newton times a meter is a joule. Since $F(x)\ge0$ on $[0,2]$, the work is the positive shaded area under the force curve.