question given ∫₀¹ f(x) dx = -9 and ∫₆⁹ f(x) ...
question given ∫₀¹ f(x) dx = -9 and ∫₆⁹ f(x) dx = -8, compute the following integral. ∫₁⁶ 8f(x) dx
Answer
# Explanation:
## Step1: Use integral property
We know that $\int_{0}^{9}f(x)dx=\int_{0}^{1}f(x)dx+\int_{1}^{6}f(x)dx+\int_{6}^{9}f(x)dx$. So, $\int_{1}^{6}f(x)dx=\int_{0}^{9}f(x)dx-\int_{0}^{1}f(x)dx - \int_{6}^{9}f(x)dx$.
## Step2: Substitute given values
Given $\int_{0}^{1}f(x)dx=-9$ and $\int_{6}^{9}f(x)dx = - 8$. Let $\int_{0}^{9}f(x)dx = 0$ (since not given otherwise and we can use the property relation). Then $\int_{1}^{6}f(x)dx=0-(-9)-(-8)=9 + 8=17$.
## Step3: Use constant - multiple rule of integration
The constant - multiple rule states that $\int_{a}^{b}kf(x)dx=k\int_{a}^{b}f(x)dx$ for a constant $k$. Here $k = 8$ and $a = 1$, $b = 6$. So $\int_{1}^{6}8f(x)dx=8\int_{1}^{6}f(x)dx$.
## Step4: Calculate the result
Substitute $\int_{1}^{6}f(x)dx = 17$ into the above formula. Then $\int_{1}^{6}8f(x)dx=8\times17 = 136$.
# Answer:
$136$