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

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| Attempts: 12 | Questions: 15 | Updated: Feb 3, 2026
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1) Which substitution is most appropriate for ∫from 0 to 1 of (2x + 1)³ dx?

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.

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About This Quiz
Definite Integrals With U-substitution: Changing Limits & Completing The Square - Quiz

Think you can handle substitution without getting mixed up by bounds? This quiz focuses on definite integrals where you must change the limits of integration after substituting. You’ll practice converting endpoints correctly, rewriting the integral in terms of u, and evaluating answers cleanly without switching back to x.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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8) Evaluate ∫(x² - 1)^(½) dx

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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Alva Benedict B. |PhD
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Alva Benedict B. is an experienced mathematician and math content developer with over 15 years of teaching and tutoring experience across high school, undergraduate, and test prep levels. He specializes in Algebra, Calculus, and Statistics, and holds advanced academic training in Mathematics with extensive expertise in LaTeX-based math content development.
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Which substitution is most appropriate for ∫from 0 to 1 of (2x +...
Find the antiderivative of 1/(x² + 6x + 13) by completing the square
Find ∫from 0 to 2 of x(4 - x²)^(½) dx
When evaluating ∫from a to b of f(g(x))g'(x) dx using substitution,...
Evaluate ∫from 1 to e of ln(x)/x dx
Which technique correctly handles the limits of integration when using...
What is the correct approach for changing limits in substitution for...
Evaluate ∫(x² - 1)^(½) dx
Evaluate ∫(x² + 4x + 8)/(x + 2)² dx
What is the best first step for evaluating ∫(3x⁴ + 2x³ + x² +...
Compute the definite integral from 0 to π/2 of sin(x)cos(x) dx
Find the integral of (2x + 3)/(x² + 3x + 2)
Which substitution simplifies ∫1/(x² + 2x + 5) dx?
Which technique is most appropriate for ∫(x² + 4x + 5)^(½) dx?
Among the choices, which is the substitution for evaluating...
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