Question

consider two functions f and g on 3,7 such that ∫₃⁷f(x)dx = 14, ∫₃⁷g(x)dx = 7, ∫₅⁷f(x)dx = 8, and ∫₃⁵g(x)dx = 4. evaluate the fo

a. ∫₃⁵2f(x)dx = (simplify your answer.)

b. ∫₃⁷(f(x) - g(x))dx = (simplify your answer.)

c. ∫₃⁵(f(x) - g(x))dx = (simplify your answer.)

d. ∫₅⁷(g(x) - f(x))dx = (simplify your answer.)

e. ∫₅⁷7g(x)dx = (simplify your answer.)

f. ∫₇³4f(x)dx = (simplify your answer.)

Explanation:

Part a

Step1: Recall integral property

The integral of a constant multiple of a function is the constant multiple of the integral, i.e., $\int_{a}^{b} kf(x)dx = k\int_{a}^{b} f(x)dx$. Also, we know that $\int_{3}^{7}f(x)dx=\int_{3}^{5}f(x)dx+\int_{5}^{7}f(x)dx$, so $\int_{3}^{5}f(x)dx=\int_{3}^{7}f(x)dx - \int_{5}^{7}f(x)dx$. Given $\int_{3}^{7}f(x)dx = 14$ and $\int_{5}^{7}f(x)dx = 8$, then $\int_{3}^{5}f(x)dx=14 - 8 = 6$.

Step2: Apply the constant multiple rule

For $\int_{3}^{5}2f(x)dx$, using the property $\int_{a}^{b} kf(x)dx = k\int_{a}^{b} f(x)dx$ with $k = 2$ and $\int_{3}^{5}f(x)dx=6$, we get $2\times6 = 12$.

Step1: Recall integral linearity

The integral of a difference of functions is the difference of integrals, i.e., $\int_{a}^{b}(f(x)-g(x))dx=\int_{a}^{b}f(x)dx-\int_{a}^{b}g(x)dx$.

Step2: Substitute the given values

We know $\int_{3}^{7}f(x)dx = 14$ and $\int_{3}^{7}g(x)dx = 7$. So $\int_{3}^{7}(f(x)-g(x))dx=\int_{3}^{7}f(x)dx-\int_{3}^{7}g(x)dx=14 - 7 = 7$.

Step1: Recall integral linearity

$\int_{a}^{b}(f(x)-g(x))dx=\int_{a}^{b}f(x)dx-\int_{a}^{b}g(x)dx$.

Step2: Find $\int_{3}^{5}f(x)dx$ and $\int_{3}^{5}g(x)dx$

We already found $\int_{3}^{5}f(x)dx = 6$ (from part a). For $\int_{3}^{5}g(x)dx$, using $\int_{3}^{7}g(x)dx=\int_{3}^{5}g(x)dx+\int_{5}^{7}g(x)dx$, so $\int_{3}^{5}g(x)dx=\int_{3}^{7}g(x)dx-\int_{5}^{7}g(x)dx$. Given $\int_{3}^{7}g(x)dx = 7$ and $\int_{5}^{7}g(x)dx = 4$, then $\int_{3}^{5}g(x)dx=7 - 4 = 3$.

Step3: Substitute into the formula

$\int_{3}^{5}(f(x)-g(x))dx=\int_{3}^{5}f(x)dx-\int_{3}^{5}g(x)dx=6 - 3 = 3$.

Answer:

12

Part b