Answer :
Alright, let's evaluate each of the integrals step-by-step.
### Question 29
Evaluate: [tex]\(\int_1^3 x^2(4x + 3) \, dx\)[/tex].
First, simplify the integrand:
[tex]\[ x^2(4x + 3) = 4x^3 + 3x^2 \][/tex]
Now, find the antiderivative:
[tex]\[ \int (4x^3 + 3x^2) \, dx = \int 4x^3 \, dx + \int 3x^2 \, dx \][/tex]
[tex]\[ = 4 \int x^3 \, dx + 3 \int x^2 \, dx \][/tex]
[tex]\[ = 4 \left(\frac{x^4}{4}\right) + 3 \left(\frac{x^3}{3}\right) \][/tex]
[tex]\[ = x^4 + x^3 \][/tex]
Now evaluate it from [tex]\(x = 1\)[/tex] to [tex]\(x = 3\)[/tex]:
[tex]\[ \left[ x^4 + x^3 \right]_1^3 = (3^4 + 3^3) - (1^4 + 1^3) \][/tex]
[tex]\[ = (81 + 27) - (1 + 1) \][/tex]
[tex]\[ = 108 - 2 \][/tex]
[tex]\[ = 106 \][/tex]
Thus, the answer is A. [tex]\(106\)[/tex].
### Question 30
Evaluate: [tex]\(\int_4^9 \frac{dx}{\sqrt{x}}\)[/tex].
Rewrite the integrand:
[tex]\[ \frac{1}{\sqrt{x}} = x^{-1/2} \][/tex]
Now, find the antiderivative:
[tex]\[ \int x^{-1/2} \, dx = \int x^{-\frac{1}{2}} \, dx \][/tex]
Applying the power rule:
[tex]\[ \int x^{-\frac{1}{2}} \, dx = \frac{x^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} = \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 2x^{\frac{1}{2}} \][/tex]
[tex]\[ = 2\sqrt{x} \][/tex]
Now evaluate it from [tex]\(x = 4\)[/tex] to [tex]\(x = 9\)[/tex]:
[tex]\[ \left[ 2\sqrt{x} \right]_4^9 = 2\sqrt{9} - 2\sqrt{4} \][/tex]
[tex]\[ = 2(3) - 2(2) \][/tex]
[tex]\[ = 6 - 4 \][/tex]
[tex]\[ = 2 \][/tex]
Thus, the answer is B. [tex]\(2\)[/tex].
### Question 31
Given: [tex]\(\int_{-2}^3 (ct - 1) \, dt = 0\)[/tex]. Find the value of [tex]\(c\)[/tex].
First, find the antiderivative:
[tex]\[ \int (ct - 1) \, dt = c \int t \, dt - \int 1 \, dt \][/tex]
[tex]\[ = c \left(\frac{t^2}{2}\right) - t \][/tex]
[tex]\[ = \frac{ct^2}{2} - t \][/tex]
Now, evaluate it from [tex]\(t = -2\)[/tex] to [tex]\(t = 3\)[/tex]:
[tex]\[ \left[ \frac{ct^2}{2} - t \right]_{-2}^3 = \left(\frac{c(3)^2}{2} - 3\right) - \left(\frac{c(-2)^2}{2} + 2\right) \][/tex]
[tex]\[ = \left(\frac{9c}{2} - 3\right) - \left(\frac{4c}{2} + 2\right) \][/tex]
[tex]\[ = \frac{9c}{2} - 3 - 2c - 2 \][/tex]
[tex]\[ = \frac{9c}{2} - \frac{4c}{2} - 5 \][/tex]
[tex]\[ = \frac{5c}{2} - 5 \][/tex]
Set it equal to 0:
[tex]\[ \frac{5c}{2} - 5 = 0 \][/tex]
[tex]\[ \frac{5c}{2} = 5 \][/tex]
[tex]\[ 5c = 10 \][/tex]
[tex]\[ c = 2 \][/tex]
Thus, the answer is D. [tex]\(2\)[/tex].
### Question 32
Evaluate: [tex]\(\int_{-1}^1 (x^3 - x) \, dx\)[/tex].
First, find the antiderivative:
[tex]\[ \int (x^3 - x) \, dx = \int x^3 \, dx - \int x \, dx \][/tex]
[tex]\[ = \left(\frac{x^4}{4}\right) - \left(\frac{x^2}{2}\right) \][/tex]
[tex]\[ = \frac{x^4}{4} - \frac{x^2}{2} \][/tex]
Now evaluate it from [tex]\(x = -1\)[/tex] to [tex]\(x = 1\)[/tex]:
[tex]\[ \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_{-1}^1 = \left(\frac{(1)^4}{4} - \frac{(1)^2}{2}\right) - \left(\frac{(-1)^4}{4} - \frac{(-1)^2}{2}\right) \][/tex]
[tex]\[ = \left(\frac{1}{4} - \frac{1}{2}\right) - \left(\frac{1}{4} - \frac{1}{2}\right) \][/tex]
[tex]\[ = \left(\frac{1}{4} - \frac{2}{4}\right) - \left(\frac{1}{4} - \frac{2}{4}\right) \][/tex]
[tex]\[ = \left(\frac{-1}{4}\right) - \left(\frac{-1}{4}\right) \][/tex]
[tex]\[ = -\frac{1}{4} + \frac{1}{4} \][/tex]
[tex]\[ = 0 \][/tex]
Thus, the answer is C. [tex]\(0\)[/tex].
### Question 33
Evaluate: [tex]\(\int_1^2 (4x^3 - 6x^2 + 1) \, dx\)[/tex].
First, find the antiderivative:
[tex]\[ \int (4x^3 - 6x^2 + 1) \, dx = 4 \int x^3 \, dx - 6 \int x^2 \, dx + \int 1 \, dx \][/tex]
[tex]\[ = 4 \left(\frac{x^4}{4}\right) - 6 \left(\frac{x^3}{3}\right) + x \][/tex]
[tex]\[ = x^4 - 2x^3 + x \][/tex]
Now evaluate it from [tex]\(x = 1\)[/tex] to [tex]\(x = 2\)[/tex]:
[tex]\[ \left[ x^4 - 2x^3 + x \right]_1^2 = (2^4 - 2(2^3) + 2) - (1^4 - 2(1^3) + 1) \][/tex]
[tex]\[ = (16 - 16 + 2) - (1 - 2 + 1) \][/tex]
[tex]\[ = (2) - (0) \][/tex]
[tex]\[ = 2 \][/tex]
Thus, the answer is A. [tex]\(2\)[/tex].
### Question 29
Evaluate: [tex]\(\int_1^3 x^2(4x + 3) \, dx\)[/tex].
First, simplify the integrand:
[tex]\[ x^2(4x + 3) = 4x^3 + 3x^2 \][/tex]
Now, find the antiderivative:
[tex]\[ \int (4x^3 + 3x^2) \, dx = \int 4x^3 \, dx + \int 3x^2 \, dx \][/tex]
[tex]\[ = 4 \int x^3 \, dx + 3 \int x^2 \, dx \][/tex]
[tex]\[ = 4 \left(\frac{x^4}{4}\right) + 3 \left(\frac{x^3}{3}\right) \][/tex]
[tex]\[ = x^4 + x^3 \][/tex]
Now evaluate it from [tex]\(x = 1\)[/tex] to [tex]\(x = 3\)[/tex]:
[tex]\[ \left[ x^4 + x^3 \right]_1^3 = (3^4 + 3^3) - (1^4 + 1^3) \][/tex]
[tex]\[ = (81 + 27) - (1 + 1) \][/tex]
[tex]\[ = 108 - 2 \][/tex]
[tex]\[ = 106 \][/tex]
Thus, the answer is A. [tex]\(106\)[/tex].
### Question 30
Evaluate: [tex]\(\int_4^9 \frac{dx}{\sqrt{x}}\)[/tex].
Rewrite the integrand:
[tex]\[ \frac{1}{\sqrt{x}} = x^{-1/2} \][/tex]
Now, find the antiderivative:
[tex]\[ \int x^{-1/2} \, dx = \int x^{-\frac{1}{2}} \, dx \][/tex]
Applying the power rule:
[tex]\[ \int x^{-\frac{1}{2}} \, dx = \frac{x^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} = \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 2x^{\frac{1}{2}} \][/tex]
[tex]\[ = 2\sqrt{x} \][/tex]
Now evaluate it from [tex]\(x = 4\)[/tex] to [tex]\(x = 9\)[/tex]:
[tex]\[ \left[ 2\sqrt{x} \right]_4^9 = 2\sqrt{9} - 2\sqrt{4} \][/tex]
[tex]\[ = 2(3) - 2(2) \][/tex]
[tex]\[ = 6 - 4 \][/tex]
[tex]\[ = 2 \][/tex]
Thus, the answer is B. [tex]\(2\)[/tex].
### Question 31
Given: [tex]\(\int_{-2}^3 (ct - 1) \, dt = 0\)[/tex]. Find the value of [tex]\(c\)[/tex].
First, find the antiderivative:
[tex]\[ \int (ct - 1) \, dt = c \int t \, dt - \int 1 \, dt \][/tex]
[tex]\[ = c \left(\frac{t^2}{2}\right) - t \][/tex]
[tex]\[ = \frac{ct^2}{2} - t \][/tex]
Now, evaluate it from [tex]\(t = -2\)[/tex] to [tex]\(t = 3\)[/tex]:
[tex]\[ \left[ \frac{ct^2}{2} - t \right]_{-2}^3 = \left(\frac{c(3)^2}{2} - 3\right) - \left(\frac{c(-2)^2}{2} + 2\right) \][/tex]
[tex]\[ = \left(\frac{9c}{2} - 3\right) - \left(\frac{4c}{2} + 2\right) \][/tex]
[tex]\[ = \frac{9c}{2} - 3 - 2c - 2 \][/tex]
[tex]\[ = \frac{9c}{2} - \frac{4c}{2} - 5 \][/tex]
[tex]\[ = \frac{5c}{2} - 5 \][/tex]
Set it equal to 0:
[tex]\[ \frac{5c}{2} - 5 = 0 \][/tex]
[tex]\[ \frac{5c}{2} = 5 \][/tex]
[tex]\[ 5c = 10 \][/tex]
[tex]\[ c = 2 \][/tex]
Thus, the answer is D. [tex]\(2\)[/tex].
### Question 32
Evaluate: [tex]\(\int_{-1}^1 (x^3 - x) \, dx\)[/tex].
First, find the antiderivative:
[tex]\[ \int (x^3 - x) \, dx = \int x^3 \, dx - \int x \, dx \][/tex]
[tex]\[ = \left(\frac{x^4}{4}\right) - \left(\frac{x^2}{2}\right) \][/tex]
[tex]\[ = \frac{x^4}{4} - \frac{x^2}{2} \][/tex]
Now evaluate it from [tex]\(x = -1\)[/tex] to [tex]\(x = 1\)[/tex]:
[tex]\[ \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_{-1}^1 = \left(\frac{(1)^4}{4} - \frac{(1)^2}{2}\right) - \left(\frac{(-1)^4}{4} - \frac{(-1)^2}{2}\right) \][/tex]
[tex]\[ = \left(\frac{1}{4} - \frac{1}{2}\right) - \left(\frac{1}{4} - \frac{1}{2}\right) \][/tex]
[tex]\[ = \left(\frac{1}{4} - \frac{2}{4}\right) - \left(\frac{1}{4} - \frac{2}{4}\right) \][/tex]
[tex]\[ = \left(\frac{-1}{4}\right) - \left(\frac{-1}{4}\right) \][/tex]
[tex]\[ = -\frac{1}{4} + \frac{1}{4} \][/tex]
[tex]\[ = 0 \][/tex]
Thus, the answer is C. [tex]\(0\)[/tex].
### Question 33
Evaluate: [tex]\(\int_1^2 (4x^3 - 6x^2 + 1) \, dx\)[/tex].
First, find the antiderivative:
[tex]\[ \int (4x^3 - 6x^2 + 1) \, dx = 4 \int x^3 \, dx - 6 \int x^2 \, dx + \int 1 \, dx \][/tex]
[tex]\[ = 4 \left(\frac{x^4}{4}\right) - 6 \left(\frac{x^3}{3}\right) + x \][/tex]
[tex]\[ = x^4 - 2x^3 + x \][/tex]
Now evaluate it from [tex]\(x = 1\)[/tex] to [tex]\(x = 2\)[/tex]:
[tex]\[ \left[ x^4 - 2x^3 + x \right]_1^2 = (2^4 - 2(2^3) + 2) - (1^4 - 2(1^3) + 1) \][/tex]
[tex]\[ = (16 - 16 + 2) - (1 - 2 + 1) \][/tex]
[tex]\[ = (2) - (0) \][/tex]
[tex]\[ = 2 \][/tex]
Thus, the answer is A. [tex]\(2\)[/tex].
