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

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

By the Fundamental Theorem of Calculus, $\int_0^6 h'(x)\,dx = h(6)-h(0) = 7-2 = 5$.

Q: What theorem connects the integral of a derivative to function values?

The Fundamental Theorem of Calculus states that the definite integral of a derivative over an interval equals the change in the original function across the endpoints.

Question

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

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

$7-2=5$

Acemy AI · Accepted Answer

> **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

Question

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

Expert Verified Solution

Step by step

Pitfall alert

Try different conditions

Further reading

FAQ

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

What theorem connects the integral of a derivative to function values?