Indefinite Integration of Polynomial and Trigonometric Functions Alıştırma Soruları — JAMB UTME

Soru 1Soru →

If $\int (6x^2 - 4\sin(2x)) \, dx = ax^3 + b\cos(2x) + C$ , where $a$ , $b$ , and $C$ are constants, what is the value of $a + b$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The value of

a + b

is

4

.

Integrating

6x^2

gives

2x^3

, and integrating

-4\sin(2x)

gives

2\cos(2x)

. Comparing

2x^3 + 2\cos(2x) + C

to

ax^3 + b\cos(2x) + C

yields

a=2

and

b=2

, giving

a+b=4

.

Adım Adım Çözüm

1

Integrate each term of the integrand

(6x^2 - 4\sin(2x))

with respect to

x

.

\int 6x^2 \, dx = \frac{6x^3}{3} = 2x^3

, and

\int -4\sin(2x) \, dx = -4 \cdot \left(-\frac{1}{2}\cos(2x)\right) = 2\cos(2x)

.

Applying the power rule for integration

\int x^n \, dx = \frac{x^{n+1}}{n+1}

and the standard trigonometric integral formula

\int \sin(kx) \, dx = -\frac{1}{k}\cos(kx)

.

2

Combine the calculated antiderivative terms and include the constant of integration

C

.

\int (6x^2 - 4\sin(2x)) \, dx = 2x^3 + 2\cos(2x) + C

.

Summing the individual term-by-term antiderivatives produces the complete indefinite integral.

3

Compare the resulting expression with

ax^3 + b\cos(2x) + C

to determine

a

and

b

.

a = 2

and

b = 2

.

Matching corresponding coefficients of

x^3

and

\cos(2x)

.

4

Compute

a + b

.

2 + 2 = 4

.

Evaluating the sum of the extracted coefficients.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions

Tahmini Süre:1m 30s

Soru 2Soru →

What is the indefinite integral $\int (4x^3 - 6\cos(3x)) \, dx$ ?

x^4 - 2\sin(3x) + C

x^4 - 2\sin(3x)

x^4 + 2\sin(3x) + C

x^4 - 18\sin(3x) + C

Cevabı ve açıklamayı göster

Cevap: $x^4 - 2\sin(3x) + C$

Cevap

x^4 - 2\sin(3x) + C

Integrating term-by-term,

\int 4x^3 \, dx = x^4

and

\int -6\cos(3x) \, dx = -2\sin(3x)

. Summing these and including the arbitrary constant

C

produces

x^4 - 2\sin(3x) + C

.

Adım Adım Çözüm

1

Integrate the polynomial term

4x^3

using the power rule

\int x^n \, dx = \frac{x^{n+1}}{n+1}

.

\int 4x^3 \, dx = 4 \cdot \frac{x^4}{4} = x^4

The power rule for integration adds

1

to the exponent and divides by the new exponent.

2

Integrate the trigonometric term

-6\cos(3x)

using

\int \cos(kx) \, dx = \frac{1}{k}\sin(kx)

.

\int -6\cos(3x) \, dx = -6 \cdot \frac{1}{3}\sin(3x) = -2\sin(3x)

Integrating cosine gives positive sine, divided by the coefficient of

x

.

3

Combine the results and append the constant of integration

C

.

x^4 - 2\sin(3x) + C

An indefinite integral represents a family of functions and requires the constant of integration

C

.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions

Soru 3Soru →

What is the indefinite integral $\int \left( 6x^2 - \frac{4}{x^2} + 4\sin(2x)\cos(2x) \right) dx$ ?

2x^3 + \frac{4}{x} - \frac{1}{2}\cos(4x) + C

2x^3 + \frac{4}{x} - \frac{1}{2}\cos(4x)

2x^3 - \frac{4}{x} + \frac{1}{2}\cos(4x) + C

12x + \frac{8}{x^3} + 8\cos(4x) + C

Cevabı ve açıklamayı göster

Cevap: $2x^3 + \frac{4}{x} - \frac{1}{2}\cos(4x) + C$

Cevap

2x^3 + \frac{4}{x} - \frac{1}{2}\cos(4x) + C

Simplifying

4\sin(2x)\cos(2x)

to

2\sin(4x)

using double-angle identities and integrating

6x^2 - 4x^{-2} + 2\sin(4x)

term-by-term yields

2x^3 + \frac{4}{x} - \frac{1}{2}\cos(4x) + C

.

Adım Adım Çözüm

1

Use the double-angle trigonometric identity to simplify the product term.

4\sin(2x)\cos(2x) = 2(2\sin(2x)\cos(2x)) = 2\sin(4x)

Applying the double-angle identity

\sin(2\theta) = 2\sin\theta\cos\theta

transforms the product into a standard single sine term.

2

Express the reciprocal power with a negative exponent.

-\frac{4}{x^2} = -4x^{-2}

Rewriting the fraction in index form allows the power rule of integration to be applied directly.

3

Integrate each component term using standard integration formulas.

\int 6x^2 dx = 2x^3

,

\int -4x^{-2} dx = \frac{-4x^{-1}}{-1} = \frac{4}{x}

, and

\int 2\sin(4x) dx = 2\left(-\frac{1}{4}\cos(4x)\right) = -\frac{1}{2}\cos(4x)

Integrating

x^n

gives

\frac{x^{n+1}}{n+1}

and integrating

\sin(ax)

yields

-\frac{1}{a}\cos(ax)

.

4

Combine the integrated terms and append the constant of integration.

2x^3 + \frac{4}{x} - \frac{1}{2}\cos(4x) + C

Indefinite integrals require an arbitrary constant

C

to represent the entire family of antiderivatives.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions

Soru 4Soru →

What is the indefinite integral $\int (3x^2 + 4\cos x) \, dx$ ?

x^3 + 4\sin x + C

x^3 - 4\sin x + C

x^3 + 4\sin x

6x + 4\sin x + C

Cevabı ve açıklamayı göster

Cevap: $x^3 + 4\sin x + C$

Cevap

x^3 + 4\sin x + C

Integrating

3x^2

gives

x^3

via the power rule, and integrating

4\cos x

gives

4\sin x

. Combining these terms along with the required constant of integration

C

results in

x^3 + 4\sin x + C

.

Adım Adım Çözüm

1

Integrate the polynomial term

3x^2

using the power rule

\int x^n dx = \frac{x^{n+1}}{n+1}

.

\int 3x^2 dx = \frac{3x^3}{3} = x^3

The power rule increases the exponent by 1 and divides by the new exponent.

2

Integrate the trigonometric term

4\cos x

using the standard integral

\int \cos x \, dx = \sin x

.

\int 4\cos x \, dx = 4\sin x

The antiderivative of cosine is positive sine.

3

Combine the results and append the constant of integration

C

.

x^3 + 4\sin x + C

All indefinite integrals must include an arbitrary constant

C

.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions

Soru 5Soru →

Given that $F(x) = \int (3x^2 - 2\sin x) \, dx$ and $F(0) = 6$ , what is the value of the constant of integration $C$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The value of the constant of integration

C

is

4

.

Integrating

3x^2 - 2\sin x

yields

F(x) = x^3 + 2\cos x + C

. Substituting

x = 0

gives

F(0) = 2(1) + C = 2 + C

. Since

F(0) = 6

, setting

2 + C = 6

yields

C = 4

.

Adım Adım Çözüm

1

Integrate the function

f(x) = 3x^2 - 2\sin x

with respect to

x

F(x) = x^3 + 2\cos x + C

The antiderivative of

3x^2

is

x^3

and the antiderivative of

-2\sin x

is

2\cos x

.

2

Evaluate

F(0)

using the antiderivative expression

F(0) = 0^3 + 2\cos(0) + C = 2 + C

Since

\cos(0) = 1

, the term

2\cos(0)

simplifies to

2

.

3

Solve for the integration constant

C

using

F(0) = 6

C = 4

Subtracting

2

from both sides of

2 + C = 6

yields

C = 4

.

Anahtar Kavram

Indefinite integration of polynomial and trigonometric functions with initial conditions

Soru 6Soru →

What is the indefinite integral $\int \left( (3x + 1)^2 - 6\sin(3x) \right) dx$ ?

3x^3 + 3x^2 + x + 2\cos(3x) + C

3x^3 + 3x^2 + x - 2\cos(3x) + C

3x^3 + 3x^2 + x + 2\cos(3x)

3x^3 + 3x^2 + x + 6\cos(3x) + C

Cevabı ve açıklamayı göster

Cevap: $3x^3 + 3x^2 + x + 2\cos(3x) + C$

Cevap

3x^3 + 3x^2 + x + 2\cos(3x) + C

Expanding

(3x+1)^2

yields

9x^2 + 6x + 1

. Integrating

9x^2 + 6x + 1

gives

3x^3 + 3x^2 + x

. Integrating

-6\sin(3x)

gives

+2\cos(3x)

. Combining these results with the mandatory constant of integration

+ C

gives

3x^3 + 3x^2 + x + 2\cos(3x) + C

.

Adım Adım Çözüm

1

Expand the squared polynomial binomial term

(3x + 1)^2 = 9x^2 + 6x + 1

Expanding the expression allows for simple term-by-term integration using the power rule.

2

Integrate the polynomial component term by term

\int (9x^2 + 6x + 1) dx = 3x^3 + 3x^2 + x

Apply the power rule

\int x^n dx = \frac{x^{n+1}}{n+1}

to each polynomial term.

3

Integrate the trigonometric component

\int -6\sin(3x) dx = -6 \left(-\frac{1}{3}\cos(3x)\right) = +2\cos(3x)

The integral of

\sin(kx)

is

-\frac{1}{k}\cos(kx)

. Multiplying by

-6

yields

+2\cos(3x)

.

4

Combine terms and add the constant of integration

3x^3 + 3x^2 + x + 2\cos(3x) + C

Indefinite integrals must include an arbitrary constant

+ C

to account for all antiderivatives.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions

Soru 7Soru →

What is the indefinite integral $\int (3x^2 - 5\sin(5x) + 2) \, dx$ ?

x^3 + \cos(5x) + 2x + C

x^3 - \cos(5x) + 2x + C

x^3 + 25\cos(5x) + 2x + C

x^3 + \cos(5x) + 2x

Cevabı ve açıklamayı göster

Cevap: $x^3 + \cos(5x) + 2x + C$

Cevap

x^3 + \cos(5x) + 2x + C

Integrating term-by-term yields

\int 3x^2 dx = x^3

,

\int -5\sin(5x) dx = \cos(5x)

, and

\int 2 dx = 2x

. Adding the arbitrary constant

C

produces

x^3 + \cos(5x) + 2x + C

.

Adım Adım Çözüm

1

Integrate the polynomial term

3x^2

using the power rule

\int x^n dx = \frac{x^{n+1}}{n+1}

\int 3x^2 dx = 3 \cdot \frac{x^3}{3} = x^3

Applying the power rule for integration.

2

Integrate the trigonometric term

-5\sin(5x)

using

\int \sin(kx) dx = -\frac{1}{k}\cos(kx)

\int -5\sin(5x) dx = -5 \left(-\frac{1}{5}\cos(5x)\right) = \cos(5x)

Integration of the sine function reverses differentiation with a positive sign change for negative sine.

3

Integrate the constant term

2

and append the constant of integration

C

\int 2 dx = 2x

, giving total antiderivative

x^3 + \cos(5x) + 2x + C

Indefinite integrals always require an arbitrary constant of integration

C

.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions

Soru 8Soru →

What is the indefinite integral $\int \left( (3 - 2x)(4x + 1) + 8\cos(4x) \right) dx$ ?

-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C

-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x)

-\frac{8}{3}x^3 + 5x^2 + 3x - 2\sin(4x) + C

-\frac{8}{3}x^3 + 5x^2 + 3x + 32\sin(4x) + C

Cevabı ve açıklamayı göster

Cevap: $-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C$

Cevap

-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C

Expanding the product

(3 - 2x)(4x + 1)

gives

-8x^2 + 10x + 3

. Integrating term-by-term yields

\int -8x^2 dx = -\frac{8}{3}x^3

,

\int 10x dx = 5x^2

,

\int 3 dx = 3x

, and

\int 8\cos(4x) dx = \frac{8}{4}\sin(4x) = 2\sin(4x)

. Adding the integration constant

C

gives the complete result

-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C

.

Adım Adım Çözüm

1

Expand the polynomial product inside the integrand

(3 - 2x)(4x + 1) = 12x + 3 - 8x^2 - 2x = -8x^2 + 10x + 3

Expanding the expression allows term-by-term integration using standard rules.

2

Rewrite the full integrand

\int \left( -8x^2 + 10x + 3 + 8\cos(4x) \right) dx

Substitute the expanded polynomial back into the integral expression.

3

Integrate each term individually

\int -8x^2 dx = -\frac{8}{3}x^3

,

\int 10x dx = 5x^2

,

\int 3 dx = 3x

, and

\int 8\cos(4x) dx = 8 \cdot \frac{\sin(4x)}{4} = 2\sin(4x)

Apply the power rule

\int x^n dx = \frac{x^{n+1}}{n+1}

and linear trigonometric rule

\int \cos(kx) dx = \frac{\sin(kx)}{k}

.

4

Combine terms and append the constant of integration

-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C

Indefinite integration requires adding an arbitrary constant

C

.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions

Tahmini Süre:2m 0s

Soru 9Soru →

What is the indefinite integral $\int (8x^3 + 12\cos(4x) - 5) \, dx$ ?

2x^4 + 3\sin(4x) - 5x + C

2x^4 + 3\sin(4x) - 5x

2x^4 - 3\sin(4x) - 5x + C

2x^4 + 48\sin(4x) - 5x + C

Cevabı ve açıklamayı göster

Cevap: $2x^4 + 3\sin(4x) - 5x + C$

Cevap

2x^4 + 3\sin(4x) - 5x + C

Integrating term by term gives

\int 8x^3 \, dx = 2x^4

,

\int 12\cos(4x) \, dx = 3\sin(4x)

, and

\int -5 \, dx = -5x

, along with the constant of integration

C

, yielding

2x^4 + 3\sin(4x) - 5x + C

.

Adım Adım Çözüm

1

Integrate the polynomial term

8x^3

\frac{8x^{3+1}}{3+1} = \frac{8x^4}{4} = 2x^4

Apply the power rule for integration:

\int x^n \, dx = \frac{x^{n+1}}{n+1}

.

2

Integrate the trigonometric term

12\cos(4x)

12 \cdot \frac{\sin(4x)}{4} = 3\sin(4x)

Apply the standard trigonometric integral rule:

\int \cos(kx) \, dx = \frac{1}{k}\sin(kx)

.

3

Integrate the constant term

-5

-5x

The integral of a constant

k

with respect to

x

is

kx

.

4

Combine all integrated terms and add the arbitrary constant of integration

2x^4 + 3\sin(4x) - 5x + C

An indefinite integral represents a family of functions and requires the addition of

+ C

.

Anahtar Kavram

Indefinite integration of polynomial and trigonometric functions

Tahmini Süre:1m 30s

Soru 10Soru →

What is the indefinite integral $\int \left( (2x - 3)^2 + 12\sin(4x) - 6\cos(3x) \right) dx$ ?

\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C

\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x)

\frac{4}{3}x^3 - 6x^2 + 9x + 3\cos(4x) - 2\sin(3x) + C

\frac{4}{3}x^3 - 6x^2 + 9x - 48\cos(4x) + 18\sin(3x) + C

Cevabı ve açıklamayı göster

Cevap: $\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C$

Cevap

\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C

Expanding

(2x-3)^2

yields

4x^2 - 12x + 9

. Integrating term by term:

\int 4x^2 dx = \frac{4}{3}x^3

,

\int -12x dx = -6x^2

,

\int 9 dx = 9x

,

\int 12\sin(4x) dx = -3\cos(4x)

, and

\int -6\cos(3x) dx = -2\sin(3x)

. Summing these results and adding the constant of integration

C

gives

\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C

.

Adım Adım Çözüm

1

Expand the squared binomial inside the integrand.

(2x - 3)^2 = 4x^2 - 12x + 9

Expanding algebraic terms simplifies the integrand into basic power terms ready for integration.

2

Integrate the polynomial terms using the power rule

\int x^n dx = \frac{x^{n+1}}{n+1}

.

\int (4x^2 - 12x + 9) dx = \frac{4}{3}x^3 - 6x^2 + 9x

Applying the power rule to each term individually gives

\frac{4x^3}{3} - \frac{12x^2}{2} + 9x

.

3

Integrate the trigonometric terms using standard integration rules

\int \sin(kx) dx = -\frac{1}{k}\cos(kx)

and

\int \cos(kx) dx = \frac{1}{k}\sin(kx)

.

\int 12\sin(4x) dx = -\frac{12}{4}\cos(4x) = -3\cos(4x) and \int -6\cos(3x) dx = -\frac{6}{3}\sin(3x) = -2\sin(3x)

Integrating sine gives negative cosine divided by the inner linear coefficient, and integrating cosine gives sine divided by the inner linear coefficient.

4

Combine all integrated terms and append the constant of integration

C

.

\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C

Indefinite integrals require the addition of an arbitrary constant of integration

C

.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions

Soru 11Soru →

What is the result of evaluating the indefinite integral $\int \left( (2x - 1)^2 - 6\cos(2x) \right) \, dx$ ?

\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C

\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x)

\frac{4}{3}x^3 - 2x^2 + x + 3\sin(2x) + C

\frac{4}{3}x^3 - 2x^2 + x - 12\sin(2x) + C

Cevabı ve açıklamayı göster

Cevap: $\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C$

Cevap

\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C

Expanding the squared term

(2x - 1)^2

yields

4x^2 - 4x + 1

. Integrating term by term using the power rule gives

\frac{4}{3}x^3 - 2x^2 + x

. The integral of

-6\cos(2x)

is

-\frac{6}{2}\sin(2x) = -3\sin(2x)

. Adding the arbitrary constant

C

results in

\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C

.

Adım Adım Çözüm

1

Expand the squared binomial term inside the integral

(2x - 1)^2 = 4x^2 - 4x + 1

Expanding the expression allows for term-by-term integration using standard integration rules.

2

Integrate each term of the expanded polynomial

\int (4x^2 - 4x + 1) \, dx = \frac{4}{3}x^3 - 2x^2 + x

Apply the power rule for integration: \int x^n \, dx = \frac{x^{n+1}}{n+1}.

3

Integrate the trigonometric term

\int -6\cos(2x) \, dx = -6 \cdot \frac{1}{2}\sin(2x) = -3\sin(2x)

The standard integral of \cos(kx) is \frac{1}{k}\sin(kx).

4

Combine all integrated terms and append the constant of integration

\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C

Indefinite integration requires an arbitrary constant of integration C.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions

Soru 12Soru →

What is the indefinite integral $\int (6x^5 + 8\sin(4x)) \, dx$ ?

x^6 - 2\cos(4x) + C

x^6 - 2\cos(4x)

x^6 + 2\cos(4x) + C

x^6 - 8\cos(4x) + C

Cevabı ve açıklamayı göster

Cevap: $x^6 - 2\cos(4x) + C$

Cevap

x^6 - 2\cos(4x) + C

Integrating

6x^5

using the power rule gives

\frac{6x^6}{6} = x^6

. Integrating

8\sin(4x)

gives

8 \left(-\frac{\cos(4x)}{4}\right) = -2\cos(4x)

. Summing these and appending the constant of integration

C

yields

x^6 - 2\cos(4x) + C

.

Adım Adım Çözüm

1

Split the integral into two individual terms

\int (6x^5 + 8\sin(4x)) \, dx = \int 6x^5 \, dx + \int 8\sin(4x) \, dx

Linearity property of integration allows term-by-term integration.

2

Integrate the polynomial term

6x^5

\int 6x^5 \, dx = 6 \cdot \frac{x^{5+1}}{5+1} = \frac{6x^6}{6} = x^6

Apply the power rule of integration:

\int x^n \, dx = \frac{x^{n+1}}{n+1}

.

3

Integrate the trigonometric term

8\sin(4x)

\int 8\sin(4x) \, dx = 8 \cdot \left( -\frac{\cos(4x)}{4} \right) = -2\cos(4x)

Apply the standard integral rule:

\int \sin(kx) \, dx = -\frac{1}{k}\cos(kx)

.

4

Combine results and add the constant of integration

x^6 - 2\cos(4x) + C

Indefinite integration requires an arbitrary constant

C

.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions

Tahmini Süre:1m 30s

Soru 13Soru →

If $y = \int (12x^2 + 6\sin(2x)) \, dx$ and $y = 10$ when $x = 0$ , what is the value of the constant of integration $C$ ?

Cevabı ve açıklamayı göster

Cevap: 13

Cevap

The value of the constant of integration

C

is

13

.

Integrating

12x^2 + 6\sin(2x)

gives

y = 4x^3 - 3\cos(2x) + C

. Substituting

x = 0

yields

y(0) = 4(0)^3 - 3\cos(0) + C = -3 + C

. Setting

-3 + C = 10

and solving for

C

gives

C = 13

.

Adım Adım Çözüm

1

Integrate the function with respect to

x

y = 4x^3 - 3\cos(2x) + C

Applying the power rule

\int ax^n \, dx = \frac{ax^{n+1}}{n+1}

and trigonometric integration rule

\int k\sin(bx) \, dx = -\frac{k}{b}\cos(bx)

.

2

Apply the initial condition

x = 0

and

y = 10

10 = 4(0)^3 - 3\cos(0) + C

Substituting the boundary values to solve for the specific constant of integration.

3

Evaluate trigonometric function and solve for

C

C = 13

Since

\cos(0) = 1

, the equation simplifies to

10 = -3 + C

, leading directly to

C = 13

.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions with Initial Conditions

Soru 14Soru →

If $y = \int (6x - 5\sin(x)) \, dx$ and $y = 12$ when $x = 0$ , what is the value of the constant of integration $C$ ?

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

7

Integrating

6x - 5\sin(x)

yields

3x^2 + 5\cos(x) + C

. Substituting

x = 0

and

y = 12

gives

12 = 3(0)^2 + 5(1) + C = 5 + C

, which gives

C = 7

.

Adım Adım Çözüm

1

Integrate each term of

6x - 5\sin(x)

with respect to

x

y = 3x^2 + 5\cos(x) + C

The integral of

6x

is

3x^2

, and the integral of

-5\sin(x)

is

+5\cos(x)

.

2

Substitute the initial conditions

x = 0

and

y = 12

to calculate

C

12 = 3(0)^2 + 5\cos(0) + C \implies 12 = 5 + C \implies C = 7

Since

\cos(0) = 1

, the right-hand side reduces to

5 + C

.

Anahtar Kavram

Indefinite Integration and Initial Value Evaluation

Soru 15Soru →

Find the indefinite integral $\int \left( 3x(x - 2) + 8\cos(4x) \right) dx$ .

x^3 - 3x^2 + 2\sin(4x) + C

x^3 - 3x^2 + 2\sin(4x)

x^3 - 3x^2 - 2\sin(4x) + C

x^3 - 3x^2 + 32\sin(4x) + C

Cevabı ve açıklamayı göster

Cevap: $x^3 - 3x^2 + 2\sin(4x) + C$

Cevap

x^3 - 3x^2 + 2\sin(4x) + C

Expanding

3x(x - 2)

yields

3x^2 - 6x

. Integrating

3x^2

gives

x^3

, integrating

-6x

yields

-3x^2

, and integrating

8\cos(4x)

yields

2\sin(4x)

. Adding the arbitrary constant

C

produces

x^3 - 3x^2 + 2\sin(4x) + C

.

Adım Adım Çözüm

1

Expand the algebraic term inside the integrand.

3x(x - 2) = 3x^2 - 6x

, expressing the integral as

\int (3x^2 - 6x + 8\cos(4x)) dx

.

Expanding the product converts the term into standard polynomial form suitable for integration.

2

Integrate each component term using basic integration rules.

\int 3x^2 dx = x^3

,

\int -6x dx = -3x^2

, and

\int 8\cos(4x) dx = 8 \cdot \frac{\sin(4x)}{4} = 2\sin(4x)

.

Apply the power rule

\int x^n dx = \frac{x^{n+1}}{n+1}

and trigonometric rule

\int \cos(kx) dx = \frac{\sin(kx)}{k}

.

3

Combine terms and attach the constant of integration.

x^3 - 3x^2 + 2\sin(4x) + C

An indefinite integral requires an arbitrary constant

C

to represent the full set of antiderivatives.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions

Soru 16Soru →

A curve has a gradient function defined by $\frac{dy}{dx} = 12x^3 - 6\sin(3x) + 4$ . If the curve passes through the point $(0, 15)$ , what is the value of the constant of integration $C$ ?

Cevabı ve açıklamayı göster

Cevap: 13

Cevap

The constant of integration

C

is 13.

Integrating

12x^3 - 6\sin(3x) + 4

with respect to

x

yields

y = 3x^4 + 2\cos(3x) + 4x + C

. Substituting

x = 0

and

y = 15

gives

15 = 3(0)^4 + 2\cos(0) + 4(0) + C = 2 + C

, which simplifies to

C = 13

.

Adım Adım Çözüm

1

Integrate the gradient function to obtain the general equation of the curve.

y = 3x^4 + 2\cos(3x) + 4x + C

The integral of

12x^3

is

3x^4

, the integral of

-6\sin(3x)

is

+2\cos(3x)

, and the integral of

4

is

4x

.

2

Apply the initial condition

(0, 15)

to solve for

C

.

15 = 3(0)^4 + 2\cos(0) + 4(0) + C

Substitute

x = 0

and

y = 15

into the integrated equation.

3

Calculate the numerical value of

C

.

C = 13

Since

\cos(0) = 1

, the equation becomes

15 = 2 + C

, yielding

C = 13

.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions with Initial Conditions

Soru 17Soru →

If $y = \int (4x^3 - 2\sin(x)) \, dx$ and $y = 5$ when $x = 0$ , what is the value of the constant of integration $C$ ?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The constant of integration

C

is 3.

Integrating

4x^3 - 2\sin(x)

gives

y = x^4 + 2\cos(x) + C

. Substituting

x = 0

and

y = 5

yields

5 = 0 + 2(1) + C

, so

C = 3

.

Adım Adım Çözüm

1

Integrate the polynomial and trigonometric terms

y = x^4 + 2\cos(x) + C

The integral of

4x^3

with respect to

x

is

x^4

, and the integral of

-2\sin(x)

is

2\cos(x)

.

2

Apply the given initial condition

y(0) = 5

5 = 0^4 + 2\cos(0) + C \implies 5 = 2 + C

Evaluating the expression at

x = 0

uses the trigonometric property

\cos(0) = 1

.

3

Solve the linear equation for

C

C = 3

Subtracting 2 from both sides isolates

C

.

Anahtar Kavram

Determining the Constant of Integration from Initial Boundary Conditions

Soru 18Soru →

What is the indefinite integral $\int (5x^4 - 3\sin(x)) \, dx$ ?

x^5 + 3\cos(x) + C

x^5 - 3\cos(x) + C

x^5 + 3\cos(x)

20x^3 - 3\cos(x) + C

Cevabı ve açıklamayı göster

Cevap: $x^5 + 3\cos(x) + C$

Cevap

x^5 + 3\cos(x) + C

Integrating

5x^4

gives

x^5

, and integrating

-3\sin(x)

gives

+3\cos(x)

since

\int \sin(x) \, dx = -\cos(x)

. Adding the arbitrary constant of integration

C

results in

x^5 + 3\cos(x) + C

.

Adım Adım Çözüm

1

Integrate the polynomial term

5x^4

\int 5x^4 \, dx = \frac{5x^{4+1}}{4+1} = x^5

Apply the power rule for integration:

\int x^n \, dx = \frac{x^{n+1}}{n+1}

.

2

Integrate the trigonometric term

-3\sin(x)

\int -3\sin(x) \, dx = -3(-\cos(x)) = 3\cos(x)

The integral of

\sin(x)

with respect to

x

is

-\cos(x)

.

3

Combine the terms and add the constant of integration

x^5 + 3\cos(x) + C

An indefinite integral requires an arbitrary constant

C

.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions

Soru 19Soru →

A curve has a gradient function defined by $\frac{dy}{dx} = 6x^2 + 8\sin(4x) + 3$ . If the curve passes through the point $(0, 10)$ , what is the value of the constant of integration $C$ ?

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

The constant of integration

C

is

12

.

Integrating

6x^2 + 8\sin(4x) + 3

with respect to

x

gives

y = 2x^3 - 2\cos(4x) + 3x + C

. Substituting

(0, 10)

into the integrated equation gives

10 = 2(0) - 2(1) + 3(0) + C

, which leads directly to

10 = -2 + C

, so

C = 12

.

Adım Adım Çözüm

1

Integrate the gradient function to obtain the general equation of the curve.

y = 2x^3 - 2\cos(4x) + 3x + C

The antiderivative of

\sin(kx)

is

-\frac{1}{k}\cos(kx)

and the antiderivative of

x^n

is

\frac{x^{n+1}}{n+1}

.

2

Apply the given initial condition

(x, y) = (0, 10)

to solve for

C

.

10 = 2(0)^3 - 2\cos(0) + 3(0) + C \implies 10 = -2 + C

Evaluating at

x = 0

requires evaluating

\cos(0) = 1

, which leaves

-2

from the trigonometric term.

3

Solve the linear equation for

C

.

C = 12

Adding

2

to both sides of

10 = -2 + C

gives

C = 12

.

Anahtar Kavram

Indefinite Integration with Boundary Conditions

Soru 20Soru →

A curve $y = F(x)$ has a gradient function given by $\frac{dy}{dx} = 9x^2 - 8x + 6\cos(3x) - 4\sin(2x)$ . Given that $y(0) = 7$ , determine the value of the constant of integration $C$ when the antiderivative is expressed in the standard form $y = 3x^3 - 4x^2 + 2\sin(3x) + 2\cos(2x) + C$ .

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The value of the constant of integration

C

is 5.

Integrating

\frac{dy}{dx} = 9x^2 - 8x + 6\cos(3x) - 4\sin(2x)

yields

y = 3x^3 - 4x^2 + 2\sin(3x) + 2\cos(2x) + C

. Substituting

x = 0

into the expression gives

y(0) = 0 - 0 + 0 + 2(1) + C = 2 + C

. Equating to

y(0) = 7

gives

2 + C = 7

, which solves to

C = 5

.

Adım Adım Çözüm

1

Integrate the gradient function term-by-term with respect to

x

y = 3x^3 - 4x^2 + 2\sin(3x) + 2\cos(2x) + C

The antiderivative of

9x^2

is

3x^3

, of

-8x

is

-4x^2

, of

6\cos(3x)

is

2\sin(3x)

, and of

-4\sin(2x)

is

2\cos(2x)

.

2

Apply the initial boundary condition

y(0) = 7

3(0)^3 - 4(0)^2 + 2\sin(0) + 2\cos(0) + C = 7 \implies 2 + C = 7

At

x = 0

,

\sin(0) = 0

and

\cos(0) = 1

, making the non-zero constant contribution equal to

2(1) = 2

.

3

Solve for the constant of integration

C

C = 5

Subtracting 2 from both sides of

2 + C = 7

yields

C = 5

.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions with Boundary Conditions

Indefinite Integration of Polynomial and Trigonometric Functions

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