Math  /  Calculus

Question1 point)
Let 02f(x)dx=15,03f(x)dx=502g(x)dx=12,23g(x)dx=12\begin{array}{ll} \int_{0}^{2} f(x) d x=15, & \int_{0}^{3} f(x) d x=5 \\ \int_{0}^{2} g(x) d x=12, & \int_{2}^{3} g(x) d x=-12 \end{array}
Use these values to evaluate the given definite integrals. a. 02(f(x)+g(x))dx=\int_{0}^{2}(f(x)+g(x)) d x= \square b. 03(f(x)g(x))dx=\int_{0}^{3}(f(x)-g(x)) d x= \square c. 23(3f(x)+2g(x))dx=\int_{2}^{3}(3 f(x)+2 g(x)) d x= \square d. Find the value aa such that 03(af(x)+g(x))dx=0\int_{0}^{3}(a f(x)+g(x)) d x=0 a=a= \square

Studdy Solution

STEP 1

What is this asking? We're given the values of some definite integrals and we need to use them to figure out the values of other definite integrals involving sums, differences, and scalar multiples of the original functions. Watch out! Be super careful with the integral bounds!
Mixing them up is a classic mistake.
Also, remember how constants behave with integrals.

STEP 2

1. Solve integral (a)
2. Solve integral (b)
3. Solve integral (c)
4. Solve for *a*

STEP 3

Alright, let's tackle 02(f(x)+g(x))dx\int_{0}^{2}(f(x)+g(x)) dx.
Remember the **sum rule for integrals**: the integral of a sum is the sum of the integrals!

STEP 4

So, we can rewrite our integral as 02f(x)dx+02g(x)dx\int_{0}^{2} f(x) dx + \int_{0}^{2} g(x) dx.
We know both of these values! 02f(x)dx=15\int_{0}^{2} f(x) dx = \textbf{15} and 02g(x)dx=12\int_{0}^{2} g(x) dx = \textbf{12}.

STEP 5

Let's plug those values in: 15+12=27\textbf{15} + \textbf{12} = \textbf{27}.
So, 02(f(x)+g(x))dx=27\int_{0}^{2}(f(x)+g(x)) dx = \textbf{27}.
Boom!

STEP 6

Next up: 03(f(x)g(x))dx\int_{0}^{3}(f(x)-g(x)) dx.
Similar to before, we use the **difference rule**: 03f(x)dx03g(x)dx\int_{0}^{3} f(x) dx - \int_{0}^{3} g(x) dx.

STEP 7

We know 03f(x)dx=5\int_{0}^{3} f(x) dx = \textbf{5}.
But we don't directly know 03g(x)dx\int_{0}^{3} g(x) dx.
No worries!
We know 02g(x)dx=12\int_{0}^{2} g(x) dx = \textbf{12} and 23g(x)dx=-12\int_{2}^{3} g(x) dx = \textbf{-12}.
Since integrals can be split up, we have 03g(x)dx=02g(x)dx+23g(x)dx\int_{0}^{3} g(x) dx = \int_{0}^{2} g(x) dx + \int_{2}^{3} g(x) dx.

STEP 8

Substituting the known values: 03g(x)dx=12+(-12)=0\int_{0}^{3} g(x) dx = \textbf{12} + (\textbf{-12}) = \textbf{0}.
Now we can put it all together: 03(f(x)g(x))dx=50=5\int_{0}^{3}(f(x)-g(x)) dx = \textbf{5} - \textbf{0} = \textbf{5}.

STEP 9

Time for 23(3f(x)+2g(x))dx\int_{2}^{3}(3f(x)+2g(x)) dx.
Using the sum and constant multiple rules, we get 323f(x)dx+223g(x)dx3\int_{2}^{3} f(x) dx + 2\int_{2}^{3} g(x) dx.

STEP 10

We know 23g(x)dx=-12\int_{2}^{3} g(x) dx = \textbf{-12}.
To find 23f(x)dx\int_{2}^{3} f(x) dx, we use 03f(x)dx=02f(x)dx+23f(x)dx\int_{0}^{3} f(x) dx = \int_{0}^{2} f(x) dx + \int_{2}^{3} f(x) dx.
We know 03f(x)dx=5\int_{0}^{3} f(x) dx = \textbf{5} and 02f(x)dx=15\int_{0}^{2} f(x) dx = \textbf{15}, so 5=15+23f(x)dx\textbf{5} = \textbf{15} + \int_{2}^{3} f(x) dx.
This means 23f(x)dx=515=-10\int_{2}^{3} f(x) dx = \textbf{5} - \textbf{15} = \textbf{-10}.

STEP 11

Plugging everything back in: 3(-10)+2(-12)=-3024=-543 \cdot (\textbf{-10}) + 2 \cdot (\textbf{-12}) = \textbf{-30} - \textbf{24} = \textbf{-54}.
So, 23(3f(x)+2g(x))dx=-54\int_{2}^{3}(3f(x)+2g(x)) dx = \textbf{-54}.

STEP 12

Last one: 03(af(x)+g(x))dx=0\int_{0}^{3}(af(x)+g(x)) dx = 0.
Using our rules, we get a03f(x)dx+03g(x)dx=0a\int_{0}^{3} f(x) dx + \int_{0}^{3} g(x) dx = 0.

STEP 13

We already found 03f(x)dx=5\int_{0}^{3} f(x) dx = \textbf{5} and 03g(x)dx=0\int_{0}^{3} g(x) dx = \textbf{0}.
Substituting: a5+0=0a \cdot \textbf{5} + \textbf{0} = 0, which simplifies to 5a=05a = 0.

STEP 14

Dividing both sides by 5\textbf{5} gives us a=0a = \textbf{0}.
Fantastic!

STEP 15

a. 2727 b. 55 c. 54-54 d. 00

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