Math / Calculus

Question29. $\int_{1}^{9} \frac{x-1}{\sqrt{x}} d x$
30. $\int_{0}^{2}(y-1)(2 y+1) d y$
31. $\int_{0}^{\pi / 4} \sec ^{2} t d t$
32. $\int_{0}^{\pi / 4} \sec \theta \tan \theta d \theta$
33. $\int_{1}^{2}(1+2 y)^{2} d y$
34. $\int_{0}^{3}\left(2 \sin x-e^{x}\right) d x$

Studdy Solution

STEP 1

1. Simplify the integrand.
2. Integrate term by term.
3. Evaluate the definite integral.
4. 1. Expand the integrand.
2. Integrate term by term.
3. Evaluate the definite integral.
STEP_1: High_Level_Step: 1
Expand the integrand:
$(y-1)(2y+1) = 2y^2 + y - 2y - 1 = 2y^2 - y - 1$
High_Level_Step_Completed: TRUE
STEP_2: High_Level_Step: 2
Integrate term by term:
$\int (2y^2 - y - 1) \, dy = \int 2y^2 \, dy - \int y \, dy - \int 1 \, dy$
$= \frac{2}{3}y^3 - \frac{1}{2}y^2 - y$
High_Level_Step_Completed: TRUE
STEP_3: High_Level_Step: 3
Evaluate the definite integral from 0 to 2:
$\left[ \frac{2}{3}y^3 - \frac{1}{2}y^2 - y \right]_{0}^{2}$
$= \left( \frac{2}{3}(2)^3 - \frac{1}{2}(2)^2 - (2) \right) - \left( \frac{2}{3}(0)^3 - \frac{1}{2}(0)^2 - (0) \right)$
$= \left( \frac{2}{3}(8) - \frac{1}{2}(4) - 2 \right)$
$= \left( \frac{16}{3} - 2 - 2 \right)$
$= \frac{16}{3} - 4 = \frac{16}{3} - \frac{12}{3} = \frac{4}{3}$
High_Level_Step_Completed: TRUE
The value of the integral is:
$\boxed{\frac{4}{3}}$
Problem 31:
$\int_{0}^{\pi / 4} \sec^2 t \, dt$
_HIGH_LEVEL_APPROACH_:
1. Recognize the antiderivative.
2. Evaluate the definite integral.
STEP_1: High_Level_Step: 1
Recognize the antiderivative:
The antiderivative of $\sec^2 t$ is $\tan t$ .
High_Level_Step_Completed: TRUE
STEP_2: High_Level_Step: 2
Evaluate the definite integral from 0 to $\pi/4$ :
$\left[ \tan t \right]_{0}^{\pi/4}$
$= \tan\left(\frac{\pi}{4}\right) - \tan(0)$
$= 1 - 0 = 1$
High_Level_Step_Completed: TRUE
The value of the integral is:
$\boxed{1}$
Problem 32:
$\int_{0}^{\pi / 4} \sec \theta \tan \theta \, d\theta$
_HIGH_LEVEL_APPROACH_:
1. Recognize the antiderivative.
2. Evaluate the definite integral.
STEP_1: High_Level_Step: 1
Recognize the antiderivative:
The antiderivative of $\sec \theta \tan \theta$ is $\sec \theta$ .
High_Level_Step_Completed: TRUE
STEP_2: High_Level_Step: 2
Evaluate the definite integral from 0 to $\pi/4$ :
$\left[ \sec \theta \right]_{0}^{\pi/4}$
$= \sec\left(\frac{\pi}{4}\right) - \sec(0)$
$= \sqrt{2} - 1$
High_Level_Step_Completed: TRUE
The value of the integral is:
$\boxed{\sqrt{2} - 1}$
Problem 33:
$\int_{1}^{2}(1+2y)^{2} \, dy$
_HIGH_LEVEL_APPROACH_:
1. Expand the integrand.
2. Integrate term by term.
3. Evaluate the definite integral.
STEP_1: High_Level_Step: 1
Expand the integrand:
$(1+2y)^{2} = 1 + 4y + 4y^2$
High_Level_Step_Completed: TRUE
STEP_2: High_Level_Step: 2
Integrate term by term:
$\int (1 + 4y + 4y^2) \, dy = \int 1 \, dy + \int 4y \, dy + \int 4y^2 \, dy$
$= y + 2y^2 + \frac{4}{3}y^3$
High_Level_Step_Completed: TRUE
STEP_3: High_Level_Step: 3
Evaluate the definite integral from 1 to 2:
$\left[ y + 2y^2 + \frac{4}{3}y^3 \right]_{1}^{2}$
$= \left( 2 + 2(2)^2 + \frac{4}{3}(2)^3 \right) - \left( 1 + 2(1)^2 + \frac{4}{3}(1)^3 \right)$
$= \left( 2 + 8 + \frac{32}{3} \right) - \left( 1 + 2 + \frac{4}{3} \right)$
$= \left( 10 + \frac{32}{3} \right) - \left( 3 + \frac{4}{3} \right)$
$= \frac{30}{3} + \frac{32}{3} - \frac{9}{3} - \frac{4}{3}$
$= \frac{62}{3} - \frac{13}{3} = \frac{49}{3}$
High_Level_Step_Completed: TRUE
The value of the integral is:
$\boxed{\frac{49}{3}}$
Problem 34:
$\int_{0}^{3}\left(2 \sin x - e^{x}\right) \, dx$
_HIGH_LEVEL_APPROACH_:
1. Integrate term by term.
2. Evaluate the definite integral.
STEP_1: High_Level_Step: 1
Integrate term by term:
$\int \left(2 \sin x - e^{x}\right) \, dx = \int 2 \sin x \, dx - \int e^{x} \, dx$
The antiderivative of $2 \sin x$ is $-2 \cos x$ , and the antiderivative of $e^{x}$ is $e^{x}$ .
$= -2 \cos x - e^{x}$
High_Level_Step_Completed: TRUE
STEP_2: High_Level_Step: 2
Evaluate the definite integral from 0 to 3:
$\left[ -2 \cos x - e^{x} \right]_{0}^{3}$
$= \left( -2 \cos(3) - e^{3} \right) - \left( -2 \cos(0) - e^{0} \right)$
$= \left( -2 \cos(3) - e^{3} \right) - \left( -2(1) - 1 \right)$
$= -2 \cos(3) - e^{3} + 2 + 1$
$= -2 \cos(3) - e^{3} + 3$
High_Level_Step_Completed: TRUE
The value of the integral is:
$\boxed{-2 \cos(3) - e^{3} + 3}$

STEP 2

Simplify the integrand:
$\frac{x-1}{\sqrt{x}} = \frac{x}{\sqrt{x}} - \frac{1}{\sqrt{x}} = x^{1/2} - x^{-1/2}$

STEP 3

Integrate term by term:
$\int (x^{1/2} - x^{-1/2}) \, dx = \int x^{1/2} \, dx - \int x^{-1/2} \, dx$
$= \frac{x^{3/2}}{3/2} - \frac{x^{1/2}}{1/2} = \frac{2}{3}x^{3/2} - 2x^{1/2}$

STEP 4

Evaluate the definite integral from 1 to 9:
$\left[ \frac{2}{3}x^{3/2} - 2x^{1/2} \right]_{1}^{9}$
$= \left( \frac{2}{3}(9)^{3/2} - 2(9)^{1/2} \right) - \left( \frac{2}{3}(1)^{3/2} - 2(1)^{1/2} \right)$
$= \left( \frac{2}{3}(27) - 2(3) \right) - \left( \frac{2}{3}(1) - 2(1) \right)$
$= \left( 18 - 6 \right) - \left( \frac{2}{3} - 2 \right)$
$= 12 - \left( \frac{2}{3} - \frac{6}{3} \right)$
$= 12 - \left( -\frac{4}{3} \right)$
$= 12 + \frac{4}{3} = \frac{36}{3} + \frac{4}{3} = \frac{40}{3}$
The value of the integral is:
$\boxed{\frac{40}{3}}$
Problem 30:
$\int_{0}^{2}(y-1)(2y+1) \, dy$

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