Evaluate $\int_0^4 (6x^2-16x+2)\,dx$

Key concept: For a polynomial integrand, integrate each term separately, then substitute the upper and lower limits and subtract.

Step by step

Integrate term by term:

$\int (6x^2-16x+2)\,dx = 2x^3-8x^2+2x.$

Now evaluate from $0$ to $4$:

$\int_0^4 (6x^2-16x+2)\,dx = \left[2x^3-8x^2+2x\right]_0^4.$

Substitute the bounds:

$\left(2(4)^3-8(4)^2+2(4)\right)-\left(2(0)^3-8(0)^2+2(0)\right)$

$=(128-128+8)-0=8.$

Answer: $8$

Pitfall alert

A frequent error is dropping a term when integrating, especially the constant $2$. Another common mistake is forgetting to evaluate both limits and subtract the lower value.

Try different conditions

If the upper limit changed, you would plug that new value into $2x^3-8x^2+2x$. If the integrand had one extra power of $x$, each term would still be handled separately using the power rule.

Further reading

definite integral, power rule, polynomial antiderivative