Question

A force \( (3x^2 + 2x - 5) \, \text{N} \) displaces a body from \( x = 2 \, \text{m} \) to \( x = 4 \, \text{m} \). The work done by this force is _________ J.

Answer 1

Correct Answer: 58

Step 1 — Write the integral for work

The work done by a force that varies with position is the definite integral of the force over the path:

\[ W=\int_{x=2}^{x=4} \big(3x^{2}+2x-5\big)\,dx. \]

Step 2 — Find an antiderivative (indefinite integral)

Integrate term-by-term:

• \(\displaystyle \int 3x^{2}\,dx = x^{3}\)
• \(\displaystyle \int 2x\,dx = x^{2}\)
• \(\displaystyle \int (-5)\,dx = -5x\)

So an antiderivative is

\(F(x)=x^{3} + x^{2} - 5x\)

Step 3 — Evaluate the definite integral

Compute \(F(4)\) and \(F(2)\):

\[ \begin{aligned} F(4) &= 4^{3} + 4^{2} - 5\cdot 4 = 64 + 16 - 20 = 60,\\[6pt] F(2) &= 2^{3} + 2^{2} - 5\cdot 2 = 8 + 4 - 10 = 2. \end{aligned} \]

Now subtract:

\[ W = F(4) - F(2) = 60 - 2 = 58 \]

So the work done is 58 J.

Units & physical meaning

Force is in newtons (N), displacement in metres (m). The result of integration is in joules (J).

Work done = 58 J.

This positive value means the force does net positive work on the object from 2 m to 4 m.

Optional check — average force

The average force is

\(\overline{F} = \dfrac{1}{4-2}\int_{2}^{4} F(x)\,dx = \dfrac{58}{2} = 29\ \text{N}\).

Multiplying by displacement gives \(29 \times 2 = 58\) J, consistent with the integral.

Common mistakes to avoid

• Forgetting to evaluate both limits (upper minus lower).
• Incorrect signs in the antiderivative.
• Omitting units — always state joules (J).

Quick recap

Work done \(=\displaystyle\int_{2}^{4} (3x^{2}+2x-5)\,dx = \boxed{58\ \text{J}}\).