Question

$$\int\limits_{ - 2}^0 {\left\{ {{x^3} + 3{x^2} + 3x + 3 + \left( {x + 1} \right)\,\cos \left( {x + 1} \right)} \right\}dx} $$          is equal to-

A. $$ - 4$$
B. $$0$$
C. $$4$$  
D. $$6$$
Answer :   $$4$$
Solution :
$$\eqalign{ & I = \int\limits_{ - 2}^0 {\left[ {{x^3} + 3{x^2} + 3x + 3 + \left( {x + 1} \right)\,\cos \left( {x + 1} \right)} \right]dx} \cr & = \left[ {\frac{{{x^4}}}{4} + {x^3} + \frac{{3{x^2}}}{2} + 3x + \left( {x + 1} \right)\sin \left( {x + 1} \right) + \cos \left( {x + 1} \right)} \right] \cr & = \left( {\sin \,1 + \cos \,1} \right) - \left( {4 - 8 + 6 - 6 + \sin \,1 + \cos \,1} \right) \cr & = 4 \cr} $$

Releted MCQ Question on
Calculus >> Definite Integration

Releted Question 1

The value of the definite integral $$\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} \,dx$$     is-

A. $$ - 1$$
B. $$2$$
C. $$1 + {e^{ - 1}}$$
D. none of these
Releted Question 2

Let $$a,\,b,\,c$$   be non-zero real numbers such that $$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = } \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} $$
Then the quadratic equation $$a{x^2} + bx + c = 0$$     has-

A. no root in $$\left( {0,\,2} \right)$$
B. at least one root in $$\left( {0,\,2} \right)$$
C. a double root in $$\left( {0,\,2} \right)$$
D. two imaginary roots
Releted Question 3

The value of the integral $$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,x} }}{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }}dx} $$     is-

A. $$\frac{\pi }{4}$$
B. $$\frac{\pi }{2}$$
C. $$\pi $$
D. none of these
Releted Question 4

For any integer $$n$$ the integral $$\int\limits_0^\pi {{e^{{{\cos }^2}x}}} {\cos ^3}\left( {2n + 1} \right)xdx$$     has the value-

A. $$\pi $$
B. $$1$$
C. $$0$$
D. none of these

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Definite Integration


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