Question

Math

Posted 3 months ago

`$\begin{array}{l}H(x)=-x-5 \\ h(x)=H^{\prime}(x) \\ \int_{-3}^{6} h(x) d x=\end{array}$`

Sign in to unlock the answer

Answer from Sia

Posted 3 months ago

Solution by Steps

step 1

Identify the function to be integrated

step 2

Set up the definite integral for the function $H(x) = -x - 5$ from $-3$ to $6$

step 3

Compute the integral $\int_{-3}^{6} (-x - 5) \, dx$

step 4

Apply the fundamental theorem of calculus to evaluate the integral

step 5

The result of the integral is the difference $H(6) - H(-3)$

step 6

Calculate $H(6) = -(6) - 5 = -11$ and $H(-3) = -(-3) - 5 = 3 - 5 = -2$

step 7

Find the difference $H(6) - H(-3) = -11 - (-2) = -11 + 2 = -9$

step 8

The definite integral $\int_{-3}^{6} h(x) \, dx$ is equal to $-9$

Answer

$\int_{-3}^{6} h(x) \, dx = -9$

Key Concept

Definite Integral Evaluation

Explanation

The definite integral of a function over an interval is the net area under the curve, which can be found by evaluating the antiderivative at the upper limit and subtracting the antiderivative evaluated at the lower limit.

Not the question you are looking for? Ask here!

Enter question by text

Enter question by image

Upgrade to Asksia Pro

Join a AskSia's Pro Plan, and get 24/7 AI tutoring for your reviews, assignments, quizzes and exam preps.

Unlimited chat query usages

Strong algorithms that better know you

Early access to new release features

Study Other Question