4.3 MCQs-Indefinite and Definite Integration

Indefinite and Definite Integration MCQs

Antiderivatives and Indefinite Integrals

1. The indefinite integral $\int f(x) \, dx$ represents:

1. The area under the curve $f(x)$

2. The derivative of $f(x)$

3. The family of all antiderivatives of $f(x)$

4. The slope of the tangent line to $f(x)$

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Answer: 3. The family of all antiderivatives of $f(x)$

Explanation:

• If $F'(x) = f(x)$, then $F(x)$ is an antiderivative of $f(x)$.

• The indefinite integral $\int f(x) \, dx = F(x) + C$, where C is the constant of integration, represents the entire family of antiderivatives.

• Each member of this family differs by a constant.

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2. The Power Rule for integration states that for $n \neq -1$, $\int x^n \, dx =$

1. $\frac{x^{n+1}}{n+1} + C$

2. $\frac{x^{n-1}}{n-1} + C$

3. $n x^{n-1} + C$

4. $x^{n+1} + C$

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Answer: 1. $\frac{x^{n+1}}{n+1} + C$

Explanation:

• This is the reverse of the Power Rule for differentiation: $\frac{d}{dx}\left( \frac{x^{n+1}}{n+1} \right) = x^n$.

• The condition $n \neq -1$ is important because it leads to the integral $\int x^{-1} dx = \int \frac{1}{x} dx = \ln|x| + C$.

• Example: $\int x^3 dx = \frac{x^{4}}{4} + C$.

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3. $\int e^x \, dx =$

1. $\ln x + C$

2. $x e^{x-1} + C$

3. $e^x + C$

4. $\frac{e^{x+1}}{x+1} + C$

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Answer: 3. $e^x + C$

Explanation:

• Since the derivative of $e^x$ is $e^x$, its antiderivative is also $e^x$, plus the constant of integration.

• This is a unique and important property of the exponential function with base $e$.

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4. $\int \frac{1}{x} \, dx =$

1. $\ln x + C$

2. $\ln |x| + C$

3. $\frac{1}{x^2} + C$

4. $x^{-1} + C$

Show me the answer

Answer: 2. $\ln |x| + C$

Explanation:

• The absolute value is crucial because the domain of $\frac{1}{x}$ excludes $x=0$, and the antiderivative $\ln x$ is only defined for $x > 0$.

• For $x < 0$, the derivative of $\ln(-x)$ is also $\frac{1}{x}$.

• Therefore, the general antiderivative is $\ln |x| + C$, valid for all $x \neq 0$.

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Basic Integration Rules and Substitution

5. The Constant Multiple Rule for integration states: $\int k \cdot f(x) \, dx =$

1. $k + \int f(x) \, dx$

2. $k \cdot \int f(x) \, dx$

3. $\int k \, dx \cdot \int f(x) \, dx$

4. $f(kx) + C$

Show me the answer

Answer: 2. $k \cdot \int f(x) \, dx$

Explanation:

• Constants can be factored out of the integral. Formally, $\int k f(x) dx = k \int f(x) dx$, where k is any constant.

• This follows directly from the linearity of the derivative: the derivative of $k F(x)$ is $k f(x)$.

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6. The Sum Rule for integration states: $\int [f(x) + g(x)] \, dx =$

1. $\int f(x) \, dx \cdot \int g(x) \, dx$

2. $\int f(x) \, dx + \int g(x) \, dx$

3. $f(x) + g(x) + C$

4. $\frac{1}{2} \int f(x) \, dx + \frac{1}{2} \int g(x) \, dx$

Show me the answer

Answer: 2. $\int f(x) \, dx + \int g(x) \, dx$

Explanation:

• The integral of a sum is the sum of the integrals. Formally, $\int [f(x) + g(x)] dx = \int f(x) dx + \int g(x) dx$.

• Combined with the Constant Multiple Rule, this makes integration a linear operation.

• Example: $\int (3x^2 + 2\cos x) dx = \int 3x^2 dx + \int 2\cos x dx = x^3 + 2\sin x + C$.

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7. The Substitution Rule (u-substitution) for integration is essentially:

1. The Power Rule in reverse

2. The Chain Rule in reverse

3. The Product Rule in reverse

4. The Quotient Rule in reverse

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Answer: 2. The Chain Rule in reverse

Explanation:

• If an integral is of the form $\int f(g(x)) g'(x) dx$, we can set $u = g(x)$, then $du = g'(x) dx$.

• The integral then simplifies to $\int f(u) du$, which is hopefully easier to evaluate.

• This method "undoes" the Chain Rule for differentiation.

• Example: $\int 2x \cos(x^2) dx$. Let $u = x^2$, $du = 2x dx$. The integral becomes $\int \cos u \, du = \sin u + C = \sin(x^2) + C$.

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Definite Integrals and The Fundamental Theorem

8. The definite integral $\int_{a}^{b} f(x) \, dx$ geometrically represents:

1. The slope of the secant line from $(a, f(a))$ to $(b, f(b))$

2. The average value of $f(x)$ on $[a, b]$

3. The net signed area between the curve $y=f(x)$, the x-axis, and the lines $x=a$ and $x=b$

4. The derivative of $f(x)$ at $x=b$

Show me the answer

Answer: 3. The net signed area between the curve $y=f(x)$, the x-axis, and the lines $x=a$ and $x=b$

Explanation:

• Area above the x-axis is counted positively.

• Area below the x-axis is counted negatively.

• The result is the net signed area.

• This is the primary geometric interpretation of the definite integral.

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9. The First Fundamental Theorem of Calculus states that if $f$ is continuous on $[a, b]$ and $F$ is an antiderivative of $f$ (i.e., $F' = f$), then:

1. $\frac{d}{dx} \int_{a}^{x} f(t) dt = f(x)$

2. $\int_{a}^{b} f(x) dx = F(b) - F(a)$

3. Both 1 and 2

4. $\int_{a}^{b} F(x) dx = f(b) - f(a)$

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Answer: 3. Both 1 and 2

Explanation:

• The First Fundamental Theorem of Calculus has two closely related parts.

• Part 1: If $g(x) = \int_{a}^{x} f(t) dt$, then $g'(x) = f(x)$. This links differentiation and integration.

• Part 2 (Evaluation Theorem): $\int_{a}^{b} f(x) dx = F(b) - F(a)$. This provides a practical way to evaluate definite integrals using antiderivatives.

• Notation: $F(b) - F(a)$ is often written as $\left[ F(x) \right]_{a}^{b}$ or $F(x) \big|_{a}^{b}$.

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10. Using the Evaluation Theorem, $\int_{1}^{3} 2x \, dx =$

1. 4

2. 8

3. 9

4. 10

Show me the answer

Answer: 2. 8

Explanation:

• Find an antiderivative: $\int 2x dx = x^2 + C$. We can use $F(x) = x^2$.

• Apply the theorem: $\int_{1}^{3} 2x dx = F(3) - F(1) = (3)^2 - (1)^2 = 9 - 1 = 8$.

• Geometrically, this is the area of a trapezoid: base from 1 to 3, with heights 2 and 6. Area = average height (4) × width (2) = 8.

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Properties of Definite Integrals

11. The property $\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx$ reflects that:

1. Reversing the limits of integration changes the sign of the integral.

2. The integral is symmetric.

3. The area is always positive.

4. The function is odd.

Show me the answer

Answer: 1. Reversing the limits of integration changes the sign of the integral.

Explanation:

• This is a definition/convention that ensures consistency with the Evaluation Theorem.

• If $F$ is an antiderivative, then $\int_{a}^{b} f(x) dx = F(b)-F(a)$.

• Conversely, $\int_{b}^{a} f(x) dx = F(a)-F(b) = -(F(b)-F(a)) = -\int_{a}^{b} f(x) dx$.

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12. The Additivity property for definite integrals states that for $a < c < b$:

1. $\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx$

2. $\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx \cdot \int_{c}^{b} f(x) dx$

3. $\int_{a}^{b} f(x) dx = \int_{a}^{b} [f(x)+g(x)] dx$

4. $\int_{a}^{b} f(x) dx = \int_{c}^{d} f(x) dx$

Show me the answer

Answer: 1. $\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx$

Explanation:

• This property is intuitive from the area interpretation: The total area from a to b is the sum of the area from a to c and the area from c to b.

• This property is very useful for splitting integrals, especially when a function is defined piecewise or has different behaviors on different intervals.

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Integration by Parts

13. The formula for Integration by Parts is derived from:

1. The Chain Rule

2. The Product Rule

3. The Quotient Rule

4. The Power Rule

Show me the answer

Answer: 2. The Product Rule

Explanation:

• Starting from the Product Rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$.

• Integrate both sides: $u(x)v(x) = \int u'(x)v(x) dx + \int u(x)v'(x) dx$.

• Rearrange: $\int u(x)v'(x) dx = u(x)v(x) - \int u'(x)v(x) dx$.

• In differential form (letting $dv = v'(x)dx$ and $du = u'(x)dx$): $\int u \, dv = uv - \int v \, du$.

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14. Integration by Parts is most useful for integrals of the form:

1. $\int \sin(x^2) dx$

2. $\int x e^x dx$

3. $\int e^{x^2} dx$

4. $\int \frac{\ln x}{x} dx$

Show me the answer

Answer: 2. $\int x e^x dx$

Explanation:

• Integration by Parts is effective for integrals that are products of different types of functions, such as polynomial × exponential ($x^n e^x$), polynomial × trigonometric ($x^n \sin x$), or logarithmic × polynomial ($\ln x \cdot x^n$).

• For $\int x e^x dx$, a good choice is: $u = x$ (so $du = dx$) and $dv = e^x dx$ (so $v = e^x$).

• Then $\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + C = e^x(x-1) + C$.

• Options 1 and 3 do not have elementary antiderivatives. Option 4 is better solved by simple substitution (let $u = \ln x$).