Class 12 Mathematics Chapter 7 Integrals

Key Concept: Basic Formula Application

c) $$\int u \, dv = uv – \int v \, du$$
[Solution Description] The integration by parts formula is a technique for integrating products of functions and is given as: $$\int u \, dv = uv – \int v \, du$$.

Key Concept: Fundamental Theorem

a) $$f(x)$$
[Solution Description] The first fundamental theorem of calculus states that if $$A(x)$$ is the area function defined by $$A(x) = \int_a^x f(t) \, dt$$, then the derivative of $$A(x)$$ with respect to $$x$$ is equal to the function itself, i.e., $$A'(x) = f(x)$$.

Key Concept: Identity Application

b) $$\frac{x}{2} – \frac{1}{4} \sin 2x + C$$
[Solution Description]
To integrate $$\sin^2 x$$, use the identity $$\sin^2 x = \frac{1 – \cos 2x}{2}$$:

$$
\int \sin^2 x \, dx = \int \frac{1 – \cos 2x}{2} \, dx
= \frac{1}{2} \left( \int 1 \, dx – \int \cos 2x \, dx \right)
$$

Calculate each part:

– $$\int 1 \, dx = x$$
– $$\int \cos 2x \, dx = \frac{1}{2}\sin 2x$$

Substitute back:
$$
\text{Result} = \frac{x}{2} – \frac{1}{4} \sin 2x + C
$$

where $$C$$ is the constant of integration.

Hence, the correct answer is $$\frac{x}{2} – \frac{1}{4} \sin 2x + C$$.

Key Concept: Complex Function Integration, Advanced Properties

a) $$-\left(\sqrt{1-x^2} – \frac{(1-x^2)^{3/2}}{3}\right) + C$$

[Solution Description]

To solve this integral, we first choose a substitution. Let $$x = \sin(\theta)$$, then $$dx = \cos(\theta) d\theta$$. Thus, $$\sqrt{1-x^2} = \sqrt{1-\sin^2(\theta)} = \cos(\theta)$$.

The integral becomes:

$$\int \frac{\sin^3(\theta)}{\cos(\theta)} \cdot \cos(\theta) \, d\theta = \int \sin^3(\theta) \, d\theta$$

Rewrite $$\sin^3(\theta)$$ as $$\sin(\theta)(1-\cos^2(\theta))$$, and let $$u = \cos(\theta)$$, thus $$du = -\sin(\theta)\, d\theta$$.

Substitute it in to get:

$$-\int (1-u^2) \, du = -\left(u – \frac{u^3}{3}\right) + C = -\left(\cos(\theta) – \frac{\cos^3(\theta)}{3}\right) + C$$

Since $$\cos(\theta) = \sqrt{1-x^2}$$, convert back to x:

$$-\left(\sqrt{1-x^2} – \frac{(1-x^2)^{3/2}}{3}\right) + C$$

This is the required antiderivative.

Key Concept: Nested Substitution

a) $$\frac{7}{64}$$
[Solution Description]
Let $$u = \cos(x)$$, then $$du = -\sin(x) \, dx$$, so $$-\sin(x) \, dx = du$$. The limits change accordingly: when $$x = 0$$, $$u = 1$$; when $$x = \pi/6$$, $$u = \sqrt{3}/2$$. The integral becomes:

$$
\int_{1}^{\sqrt{3}/2} u^3 (-du) = -\int_{1}^{\sqrt{3}/2} u^3 \, du
$$

Solving this, we have:
$$
= -\left[\frac{u^4}{4}\right]_1^{\sqrt{3}/2}
= -\left(\frac{(\sqrt{3}/2)^4}{4} – \frac{1^4}{4}\right)
= -\left(\frac{9}{64} – \frac{16}{64}\right)
= \frac{7}{64}
$$

Key Concept: Integration Process

b) $$\frac{3x^3}{3} + C$$

[Solution Description] To integrate the function $$f(x) = 3x^2$$, we apply the power rule of integration. The formula is $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$. Here, $$n=2$$. So,

$$\int 3x^2 \, dx = 3 \cdot \frac{x^{2+1}}{2+1} + C = \frac{3x^3}{3} + C = x^3 + C.$$

Key Concept: Constant of Integration

d) C

[Solution Description] When you integrate a function indefinitely, you add a constant of integration, denoted as C, because there could be multiple antiderivatives differing by a constant.

For $$\int 4x dx$$, applying basic integration gives:

$$\int 4x dx = 2x^2 + C$$

where C is the constant of integration.

Thus, the constant of integration is C.

Key Concept: Exponential Function Integration

d) $$e^x + C$$
[Solution Description] The integral of $$e^x$$ is given by the formula $$\int e^x \, dx = e^x + C$$.

Key Concept: Area Function

c) The area under the curve from a to x.
[Solution Description] The area function $$A(x) = \int_a^x f(t) \, dt$$ represents the area under the curve of the function f(t) from a to x.

Key Concept: Function Continuity

d) Function is not continuous over [-1, 2]

[Solution Description]

The second fundamental theorem requires the function $$f(x)$$ to be continuous over the interval $$[a, b]$$. Here, $$f(x) = \frac{1}{x}$$ is not continuous on $$[-1, 2]$$ because it is undefined at $$x = 0$$. This discontinuity occurs within the integration limits.

Since continuity is a necessary condition for applying the theorem, it cannot be used in this case due to the discontinuity at $$x = 0$$.

Hence, the answer is that the function is not continuous over $$[-1, 2]$$.

Key Concept: Integration by Parts

a) $$e^x(x-1) + C$$
[Solution Description] Choose $$ u = x $$ and $$ dv = e^x \, dx $$. Then $$ du = dx $$ and $$ v = e^x $$. Applying integration by parts: $$ \int x e^x \, dx = x e^x – \int e^x \, dx = x e^x – e^x + C = e^x(x-1) + C $$.

Key Concept: Properties of Indefinite Integrals

b) They are always positive
[Solution Description] Indefinite integrals have several properties: they include a constant of integration, are linear, and are used to find families of functions differing by a constant. They do not have a unique solution because of the constant of integration.

Key Concept: Function and Derivative

a) $$\sin x + C$$
[Solution Description] To find $$f(x)$$ from $$f'(x) = \cos x$$, we integrate $$\cos x$$:

$$
\int \cos x \, dx = \sin x + C
$$

Thus, $$f(x) = \sin x + C$$, where $$C$$ is the constant of integration.

Key Concept: Definite Integral Concept

b) It is the limit of a sum representing the area under a function.
[Solution Description] A definite integral is described as the difference between the values of an antiderivative evaluated at two points or as the limit of a sum that represents the area under a curve from one point to another on the x-axis.

Key Concept: Practical Application

c) Determining the position of an object
[Solution Description] One practical application of integration is determining the position of an object when given its velocity over time. This involves calculating the area under the velocity-time graph.

Key Concept: Coefficient Matching

a) $$-\frac{1}{7}$$
[Solution Description]
Multiply both sides by $$(x+2)(x-5)$$ to clear the denominators:

$$
1 = A(x-5) + B(x+2).
$$

Expand the terms:

$$
1 = Ax – 5A + Bx + 2B.
$$

Combine like terms:

$$
1 = (A+B)x – 5A + 2B.
$$

Set up the system of equations:

\begin{align*}
A + B &= 0, \\
-5A + 2B &= 1.
\end{align*}

Solving for A using the first equation $$B = -A$$ in the second equation:

$$
-5A + 2(-A) = 1 \Rightarrow -7A = 1 \Rightarrow A = -\frac{1}{7}.
$$

Key Concept: Error Identification

c) $$\int_{-1}^{1} \frac{1}{x} \, dx$$
[Solution Description]
The integral $$\int_{-1}^{1} \frac{1}{x} \, dx$$ is problematic because the function $$\frac{1}{x}$$ has a discontinuity at $$x = 0$$, making it impossible to apply the Fundamental Theorem of Calculus directly over this interval without special techniques like improper integrals.

Key Concept: Quadratic Factors

a) $$\frac{2x}{x^2+x+1}$$
[Solution Description]
The denominator $$x^2+x+1$$ cannot be factorized further over the reals. Hence, decompose as:

$$
\frac{2x}{x^2+x+1} = \frac{Ax+B}{x^2+x+1}
$$

Equating numerators:

$$
2x = Ax + B
$$

This implies $$A = 2$$ and $$B = 0$$ for the equality to hold.

Thus, the partial fraction decomposition is:

$$
\frac{2x}{x^2+x+1} = \frac{2x}{x^2+x+1}
$$

Since no change occurs, the decomposition result remains the same.

Key Concept: Basic Definition

c) The area under the curve of $$f(t)$$ from $$t = a$$ to $$t = x$$
[Solution Description] The expression $$A(x) = \int_{a}^{x} f(t) \, dt$$ represents the area under the curve of $$f(t)$$ from $$t = a$$ to $$t = x$$. It accumulates the value of $$f(t)$$ over this interval.

Key Concept: Complex Substitution

b) $$-\sqrt{1-x^2} + C$$

[Solution Description] To solve this integral, we use the substitution $$x = \sin \theta$$, which means $$dx = \cos \theta \, d\theta$$. The integrand becomes:

$$\int \frac{\sin \theta}{\sqrt{1-\sin^2 \theta}} \cos \theta \, d\theta.$$

This simplifies to:

$$\int \sin \theta \, d\theta,$$

as $$\sqrt{1-\sin^2 \theta} = \cos \theta$$. Integrating gives:

$$-\cos \theta + C.$$

Since $$x = \sin \theta$$, then $$\theta = \arcsin x$$. So the final solution is:

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

Key Concept: Fundamental Theorem I

a) $$f(x)$$
[Solution Description] According to the first fundamental theorem of calculus, $$A'(x)$$ equals $$f(x)$$ when $$A(x) = \int_a^x f(t) \, dt$$.

Key Concept: Identity Transformation

b) $$\frac{x}{2} – \frac{\sin 2x}{4} + C$$

[Solution Description]

Using the identity $$\sin^2 x = \frac{1 – \cos 2x}{2}$$, we have:

$$\int \sin^2 x \, dx = \int \frac{1 – \cos 2x}{2} \, dx = \frac{1}{2} \int (1 – \cos 2x) \, dx$$

This simplifies to:

$$\frac{1}{2} \left( \int 1 \, dx – \int \cos 2x \, dx \right)$$

$$= \frac{1}{2} \left( x – \frac{1}{2}\sin 2x \right) + C$$

Therefore, the solution is $$\frac{x}{2} – \frac{\sin 2x}{4} + C$$.

Key Concept: Application of Second Theorem

a) 6

[Solution Description] To solve this, we need an antiderivative $$F(x)$$ of $$f(x) = 3x^2 – 2x + 1$$. An antiderivative is $$F(x) = x^3 – x^2 + x$$. Applying the second fundamental theorem:

$$\int_{0}^{2} (3x^2 – 2x + 1) \, dx = F(2) – F(0)$$

Calculate $$F(2)$$:

$$F(2) = (2)^3 – (2)^2 + 2 = 8 – 4 + 2 = 6$$

Calculate $$F(0)$$:

$$F(0) = (0)^3 – (0)^2 + 0 = 0$$

Therefore:

$$\int_{0}^{2} (3x^2 – 2x + 1) \, dx = 6 – 0 = 6$$

Key Concept: Substitution with Trigonometric Identities, Complex Definite Integral

b) $$\frac{2}{15}$$
[Solution Description]
Use the identity $$\sin^2 x = 1 – \cos^2 x$$ to rewrite the integrand. Thus,

$$
\int_0^{\pi/2} \sin^3 x \cos^2 x \, dx = \int_0^{\pi/2} (1 – \cos^2 x)\sin x \cos^2 x \, dx
$$

Let $$u = \cos x$$, then $$du = -\sin x \, dx$$, changing limits: when $$x = 0$$, $$u = 1$$; when $$x = \pi/2$$, $$u = 0$$. Substitute:

$$
-\int_1^0 (1-u^2) u^2 \, du = \int_0^1 (u^2 – u^4) \, du
$$

Integrate: $$\left[\frac{u^3}{3} – \frac{u^5}{5}\right]_0^1 = \frac{1}{3} – \frac{1}{5} = \frac{2}{15}$$.

Thus, the integral evaluates to $$\frac{2}{15}$$.

Key Concept: Simple Integration

c) $$\frac{1}{3}(\ln|x-2| – \ln|x+1|) + C$$
[Solution Description]
First, perform partial fraction decomposition:
$$
\frac{1}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}
$$
Multiply through by $$(x+1)(x-2)$$:
$$
1 = A(x-2) + B(x+1)
$$
Expand:
$$
1 = Ax – 2A + Bx + B
$$
Combine terms:
$$
1 = (A+B)x + (-2A+B)
$$
Equate coefficients:
\begin{align*}
A + B &= 0 \\
-2A + B &= 1
\end{align*}
From $$A + B = 0$$, $$B = -A$$. Substitute into the second equation:
$$
-2A – A = 1 \implies -3A = 1 \implies A = -\frac{1}{3}
$$
Therefore, $$B = \frac{1}{3}.$$
The integral becomes:
$$
\int \left( -\frac{1}{3(x+1)} + \frac{1}{3(x-2)} \right) dx
$$
Integrate each term:
$$
-\frac{1}{3}\int \frac{1}{x+1}dx + \frac{1}{3}\int \frac{1}{x-2}dx
$$
Giving us:
$$
-\frac{1}{3}\ln|x+1| + \frac{1}{3}\ln|x-2| + C
$$
Which simplifies to:
$$
\frac{1}{3}\left(\ln|x-2| – \ln|x+1|\right) + C
$$

Key Concept: Function Identification

a) $$2x^2 – 3x + 1$$

[Solution Description] By the First Fundamental Theorem of Integral Calculus, $$A'(x) = f(x)$$. Given that $$A'(x) = 2x^2 – 3x + 1$$, it follows directly that $$f(x) = 2x^2 – 3x + 1$$.

Key Concept: Second Function Integration

a) $$\sin x$$
[Solution Description] When choosing $$\cos x$$ as the second function, its integral is calculated. The integral of $$\cos x$$ with respect to x is $$\sin x$$. Thus, $$\int \cos x \, dx = \sin x.$$