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lakkis [162]
3 years ago
12

AP Calc BC help!

Mathematics
1 answer:
Brrunno [24]3 years ago
4 0

Answer:

Question 1) 72

Question 2) 40

Question 3) Both C and D.

Question 4) Both A and C.

Step-by-step explanation:

Question 1)

We are given that:

\displaystyle \int_1^5f(x)\, dx=8\text{ and we want to find} \int_1^5xf'(x)\, dx

We will use integration by parts, given by:

\displaystyle \int_a^b u\, dv=uv-\int_a^b v\, du

We will let:

u=x\Rightarrow du=dx\text{ and } dv=f'(x)\, dx \Rightarrow v=f(x)

Therefore:

\displaystyle \int_1^5xf'(x)\, dx=xf(x)\Big|_1^5-\int_1^5f(x)\, dx

Substitute:

\displaystyle \int_1^5xf'(x)\, dx=(5(f(5))-(1(f(1))-(8)

Evaluate:

\displaystyle \int_1^5xf'(x)\, dx=75-(-5)-8=72

Question 2)

Similarly, we will let:

\displaystyle u=x\Rightarrow du=dx\text{ and } dv=f'(x)\, dx \text{ so } v=f(x)

Hence:

\displaystyle \int_0^3 xf'(x)\, dx=xf(x)\Big|_0^3-\int_0^3f(x)\, dx

Evaluate:

\displaystyle \int_0^3 xf'(x)\, dx=(3f(3))-(0(f(0))-(2)

Thus:

\displaystyle \int_0^3 xf'(x)\, dx=3(14)-2=40

Question 3)

We are given:

\displaystyle g(x)=\int_4^xf(x)\, dx

By the Fundamental Theorem of Calculus:

g'(x)=f(x)>0

The derivative of g is always positive. So, the values of g are always increasing.

The tables that reflect this are C and D.

And there are, as I understand it, no way to determine their exact values. Both C and D are correct.

Question 4)

Similarly, we are given:

\displaystyle g(x)=\int_{-2}^xf(x)\, dx

By the FTC:

g'(x)=f(x)

So, g should be decreasing for all x.

The tables that reflect this are A and C.

So, both A and C are correct.

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