Evaluate $\int_0^6 h'(x)\,dx$

Key concept: This problem uses the Fundamental Theorem of Calculus: the integral of a derivative over an interval equals the change in the original function over that interval.

Step by step

By the Fundamental Theorem of Calculus,

$\int_0^6 h'(x)\,dx = h(6)-h(0).$

From the values given in the problem:

$h(6)=7, \quad h(0)=2.$

So,

$\int_0^6 h'(x)\,dx = 7-2=5.$

Answer: $5$

Pitfall alert

Do not try to integrate $h'(x)$ without using the endpoint values if the problem already gives $h(0)$ and $h(6)$. For a derivative, the quickest route is usually the Fundamental Theorem of Calculus.

Try different conditions

If the limits were reversed, for example $\int_6^0 h'(x)\,dx$, the answer would be the negative of this result: $-5$. If the interval changed, you would use the new endpoint values in $h(b)-h(a)$.

Further reading

Fundamental Theorem of Calculus, derivative integral, endpoint values