Definite Integrals With U-Substitution: Changing Limits & Completing The Square

1) Which substitution is most appropriate for ∫from 0 to 1 of (2x + 1)³ dx?

U = (2x + 1)³

U = 2x + 1

U = x

U = 2x

When we have a polynomial function of a linear expression, we should substitute for the linear expression to simplify the integral. Letting u = 2x + 1 makes the integral straightforward because du = 2 dx, which allows us to substitute directly. The limits also change easily: when x = 0, u = 1; when x = 1, u = 3. This gives us ∫from u=1 to u=3 of u³(du/2) = (½)[u⁴/4]from 1 to 3 = (1/8)[81 - 1] = 80/8 = 10.

Explanation

When we have a polynomial function of a linear expression, we should substitute for the linear expression to simplify the integral. Letting u = 2x + 1 makes the integral straightforward because du = 2 dx, which allows us to substitute directly. The limits also change easily: when x = 0, u = 1; when x = 1, u = 3. This gives us ∫from u=1 to u=3 of u³(du/2) = (½)[u⁴/4]from 1 to 3 = (1/8)[81 - 1] = 80/8 = 10.

2) Find the antiderivative of 1/(x² + 6x + 13) by completing the square

(1/2)arctan((x + 3)/2) + C

Arctan((x + 3)/2) + C

Arctan(x + 3) + C

Arctan(x/2) + C

First, complete the square in the denominator: x² + 6x + 13 = (x² + 6x + 9) + 4 = (x + 3)² + 4. So the integral is ∫1/((x + 3)² + 4) dx. Let u = x + 3, then du = dx. The integral becomes ∫1/(u² + 4) du = ∫1/(u² + 2²) du. The standard form is ∫1/(u² + a²) du = (1/a)arctan(u/a) + C. Here a = 2, so ∫1/(u² + 4) du = (½)arctan(u/2) + C = (½)arctan((x + 3)/2) + C.

Explanation

First, complete the square in the denominator: x² + 6x + 13 = (x² + 6x + 9) + 4 = (x + 3)² + 4. So the integral is ∫1/((x + 3)² + 4) dx. Let u = x + 3, then du = dx. The integral becomes ∫1/(u² + 4) du = ∫1/(u² + 2²) du. The standard form is ∫1/(u² + a²) du = (1/a)arctan(u/a) + C. Here a = 2, so ∫1/(u² + 4) du = (½)arctan(u/2) + C = (½)arctan((x + 3)/2) + C.

3) Find ∫from 0 to 2 of x(4 - x²)^(½) dx

8/3

16/3

4/3

2/3

Let u = 4 - x². Then du = -2x dx, so x dx = -du/2. When x = 0, u = 4. When x = 2, u = 0. Substituting, the integral becomes ∫from u=4 to u=0 of √u(-du/2) = -∫from 4 to 0 of u^1/2(du/2) = -1/2 ∫from 4 to 0 of u^1/2 du = -1/2[-∫from 0 to 4 of u^1/2 du] = 1/2 ∫from 0 to 4 of u^1/2 du. Integrating from 0 to 4: ∫u^1/2 du from 0 to 4 = [2u³/2/3]from 0 to 4 = 2/3(4³/2 - 0) = 2/3(8) = 16/3. Then multiply by 1/2: (½)(16/3) = 8/3.

Explanation

Let u = 4 - x². Then du = -2x dx, so x dx = -du/2. When x = 0, u = 4. When x = 2, u = 0. Substituting, the integral becomes ∫from u=4 to u=0 of √u(-du/2) = -∫from 4 to 0 of u^1/2(du/2) = -1/2 ∫from 4 to 0 of u^1/2 du = -1/2[-∫from 0 to 4 of u^1/2 du] = 1/2 ∫from 0 to 4 of u^1/2 du. Integrating from 0 to 4: ∫u^1/2 du from 0 to 4 = [2u³/2/3]from 0 to 4 = 2/3(4³/2 - 0) = 2/3(8) = 16/3. Then multiply by 1/2: (½)(16/3) = 8/3.

4) When evaluating ∫from a to b of f(g(x))g'(x) dx using substitution, what are the new limits?

From g(a) to g(b)

From g(b) to g(a)

From a to b

From 0 to 1

When we let u = g(x), then du = g'(x) dx. When x = a, u = g(a), and when x = b, u = g(b). Therefore, the new limits of integration are from u = g(a) to u = g(b). This is because we are changing the variable of integration from x to u, and we must use the corresponding values of the new variable u at the original limits of the old variable x.

Explanation

When we let u = g(x), then du = g'(x) dx. When x = a, u = g(a), and when x = b, u = g(b). Therefore, the new limits of integration are from u = g(a) to u = g(b). This is because we are changing the variable of integration from x to u, and we must use the corresponding values of the new variable u at the original limits of the old variable x.

5) Evaluate ∫from 1 to e of ln(x)/x dx

1/2

1

1/3

2

Let u = ln(x). Then du = (1/x) dx. When x = 1, u = 0. When x = e, u = 1. Substituting, the integral becomes ∫from u=0 to u=1 of u du = [u²/2]from 0 to 1 = 1/2 - 0 = 1/2.

Explanation

Let u = ln(x). Then du = (1/x) dx. When x = 1, u = 0. When x = e, u = 1. Substituting, the integral becomes ∫from u=0 to u=1 of u du = [u²/2]from 0 to 1 = 1/2 - 0 = 1/2.

6) Which technique correctly handles the limits of integration when using substitution?

Keep the original limits and change the variable

Change the limits to match the new variable

Only change the upper limit

Only change the lower limit

When using substitution for definite integrals, we must change both the variable of integration and the limits of integration to maintain the same value for the integral. The new limits correspond to the old limits substituted into the substitution equation. This ensures that we are evaluating the same area under the curve, just with respect to the new variable.

Explanation

When using substitution for definite integrals, we must change both the variable of integration and the limits of integration to maintain the same value for the integral. The new limits correspond to the old limits substituted into the substitution equation. This ensures that we are evaluating the same area under the curve, just with respect to the new variable.

7) Compute the definite integral from 0 to π/2 of sin(x)cos(x) dx

1/2

1/4

1/3

1/6

Let u = sin(x). Then du = cos(x) dx. When x = 0, u = 0. When x = π/2, u = 1. Substituting, the integral becomes ∫from u=0 to u=1 of u du = [u²/2]from 0 to 1 = 1/2 - 0 = 1/2.

Explanation

Let u = sin(x). Then du = cos(x) dx. When x = 0, u = 0. When x = π/2, u = 1. Substituting, the integral becomes ∫from u=0 to u=1 of u du = [u²/2]from 0 to 1 = 1/2 - 0 = 1/2.

8) What is the correct approach for changing limits in substitution for definite integrals?

The limits are always from 0 to 1

The limits are multiplied by the constant in the substitution

The limits are evaluated using the substitution equation

The limits are divided by the constant in the substitution

When we substitute u = g(x), we need to find the corresponding values of u when x equals the original limits. Specifically, if the original limits are x = a and x = b, then the new limits are u = g(a) and u = g(b). This is done by substituting the original limit values into the substitution equation to find the corresponding values of the new variable u.

Explanation

When we substitute u = g(x), we need to find the corresponding values of u when x equals the original limits. Specifically, if the original limits are x = a and x = b, then the new limits are u = g(a) and u = g(b). This is done by substituting the original limit values into the substitution equation to find the corresponding values of the new variable u.

9) Which technique is most appropriate for ∫(x² + 4x + 5)^(½) dx?

Direct substitution

Trigonometric substitution

Integration by parts

Long division

The expression x² + 4x + 5 can be completed to (x + 2)² + 1, which is of the form u² + a². Integrals of the form ∫√(u² + a²) du are best handled using trigonometric substitution, specifically u = a tan(θ). This gives √(u² + a²) = √(a² tan²(θ) + a²) = a√(tan²(θ) + 1) = a sec(θ), which simplifies the integral.

Explanation

The expression x² + 4x + 5 can be completed to (x + 2)² + 1, which is of the form u² + a². Integrals of the form ∫√(u² + a²) du are best handled using trigonometric substitution, specifically u = a tan(θ). This gives √(u² + a²) = √(a² tan²(θ) + a²) = a√(tan²(θ) + 1) = a sec(θ), which simplifies the integral.

10) Among the choices, which is the substitution for evaluating ∫1/(x√(x² + 1)) dx?

U = √(x² + 1)

U = x² + 1

U = x

U = 1/x

Let u = 1/x, so x = 1/u and dx = -1/u² du. The integral becomes ∫ (u / √((1/u²) + 1)) * (-1/u²) du = ∫ (u / √( (1+u²)/u² )) * (-1/u²) du. Assuming x > 0 (so u > 0), this simplifies to ∫ (u / (√(1+u²)/u)) * (-1/u²) du = ∫ (u² / √(1+u²)) * (-1/u²) du = ∫ -1/√(1+u²) du. This is a standard integral form, resulting in -ln|u + √(1+u²)| + C. Substituting back u = 1/x gives -ln|(1/x) + √(1+(1/x)²)| + C.

Explanation

Let u = 1/x, so x = 1/u and dx = -1/u² du. The integral becomes ∫ (u / √((1/u²) + 1)) * (-1/u²) du = ∫ (u / √( (1+u²)/u² )) * (-1/u²) du. Assuming x > 0 (so u > 0), this simplifies to ∫ (u / (√(1+u²)/u)) * (-1/u²) du = ∫ (u² / √(1+u²)) * (-1/u²) du = ∫ -1/√(1+u²) du. This is a standard integral form, resulting in -ln|u + √(1+u²)| + C. Substituting back u = 1/x gives -ln|(1/x) + √(1+(1/x)²)| + C.

11) Evaluate ∫(x² - 1)^(½) dx

(x√(x² - 1) - ln|x + √(x² - 1)|)/2 + C

(x√(x² - 1))/2 + C

X√(x² - 1) - ln|x + √(x² - 1)| + C

√(x² - 1)/2 + C

This is a standard integral that requires trigonometric substitution. Let x = sec(θ), then dx = sec(θ)tan(θ) dθ, and √(x² - 1) = √(sec²(θ) - 1) = tan(θ) (for θ in the appropriate range). The integral becomes ∫tan(θ)sec(θ)tan(θ) dθ = ∫sec(θ)tan²(θ) dθ = ∫sec(θ)(sec²(θ) - 1) dθ = ∫(sec³(θ) - sec(θ)) dθ. This requires integration by parts, but the result is (x√(x² - 1) - ln|x + √(x² - 1)|)/2 + C.

Explanation

This is a standard integral that requires trigonometric substitution. Let x = sec(θ), then dx = sec(θ)tan(θ) dθ, and √(x² - 1) = √(sec²(θ) - 1) = tan(θ) (for θ in the appropriate range). The integral becomes ∫tan(θ)sec(θ)tan(θ) dθ = ∫sec(θ)tan²(θ) dθ = ∫sec(θ)(sec²(θ) - 1) dθ = ∫(sec³(θ) - sec(θ)) dθ. This requires integration by parts, but the result is (x√(x² - 1) - ln|x + √(x² - 1)|)/2 + C.

12) Find the integral of (2x + 3)/(x² + 3x + 2)

(1/2)ln|x² + 3x + 2| + C

Ln|x + 1| + ln|x + 2| + C

Ln|x + 1| - ln|x + 2| + C

2ln|x + 1| + C

Notice that the numerator 2x + 3 is almost the derivative of the denominator x² + 3x + 2. Indeed, the derivative of x² + 3x + 2 is 2x + 3, which matches the numerator. Therefore, ∫(2x + 3)/(x² + 3x + 2) dx = ln|x² + 3x + 2| + C. But we can also write this as ln|(x + 1)(x + 2)| + C = ln|x + 1| + ln|x + 2| + C, which is option B.

Explanation

Notice that the numerator 2x + 3 is almost the derivative of the denominator x² + 3x + 2. Indeed, the derivative of x² + 3x + 2 is 2x + 3, which matches the numerator. Therefore, ∫(2x + 3)/(x² + 3x + 2) dx = ln|x² + 3x + 2| + C. But we can also write this as ln|(x + 1)(x + 2)| + C = ln|x + 1| + ln|x + 2| + C, which is option B.

13) Evaluate ∫(x² + 4x + 8)/(x + 2)² dx

X - 4/(x + 2) + C

X + 2 + 4/(x + 2) + C

X - 2/(x + 2) + C

X + 4/(x + 2) + C

Notice that the denominator is (x + 2)², and the numerator is x² + 4x + 8 = (x² + 4x + 4) + 4 = (x + 2)² + 4. So we can write the integrand as [(x + 2)² + 4]/(x + 2)² = 1 + 4/(x + 2)². Therefore, the integral is ∫(1 + 4/(x + 2)²) dx = x + 4∫(x + 2)^(-2) dx = x + 4(-(x + 2)^(-1)) + C = x - 4/(x + 2) + C.

Explanation

Notice that the denominator is (x + 2)², and the numerator is x² + 4x + 8 = (x² + 4x + 4) + 4 = (x + 2)² + 4. So we can write the integrand as [(x + 2)² + 4]/(x + 2)² = 1 + 4/(x + 2)². Therefore, the integral is ∫(1 + 4/(x + 2)²) dx = x + 4∫(x + 2)^(-2) dx = x + 4(-(x + 2)^(-1)) + C = x - 4/(x + 2) + C.

14) Which substitution simplifies ∫1/(x² + 2x + 5) dx?

U = x² + 2x + 5

U = x + 1

U = x

U = 2x + 2

First, complete the square: x² + 2x + 5 = (x² + 2x + 1) + 4 = (x + 1)² + 4. So the integral is ∫1/((x + 1)² + 4) dx. Let u = x + 1, then du = dx. The integral becomes ∫1/(u² + 4) du = (½)arctan(u/2) + C = (½)arctan((x + 1)/2) + C. The substitution u = x + 1 directly leads to the standard form ∫1/(u² + a²) du, which has a known antiderivative.

Explanation

First, complete the square: x² + 2x + 5 = (x² + 2x + 1) + 4 = (x + 1)² + 4. So the integral is ∫1/((x + 1)² + 4) dx. Let u = x + 1, then du = dx. The integral becomes ∫1/(u² + 4) du = (½)arctan(u/2) + C = (½)arctan((x + 1)/2) + C. The substitution u = x + 1 directly leads to the standard form ∫1/(u² + a²) du, which has a known antiderivative.

15) What is the best first step for evaluating ∫(3x⁴ + 2x³ + x² + 1)/(x² + 1) dx?

Substitute u = x² + 1

Use trigonometric substitution

Perform polynomial long division

Factor the numerator

Since the degree of the numerator (4) is greater than the degree of the denominator (2), we should perform polynomial long division first. This will give us a polynomial plus a proper rational function, which can then be integrated more easily. The division of 3x⁴ + 2x³ + x² + 1 by x² + 1 will simplify the expression and make the integration more straightforward.

Explanation

Since the degree of the numerator (4) is greater than the degree of the denominator (2), we should perform polynomial long division first. This will give us a polynomial plus a proper rational function, which can then be integrated more easily. The division of 3x⁴ + 2x³ + x² + 1 by x² + 1 will simplify the expression and make the integration more straightforward.

Definite Integrals with u-Substitution: Changing Limits & Completing the Square